math 5120: complex analysis. homework 1 solutions · math 5120: complex analysis. homework 1...
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Math 5120: Complex analysis. Homework 1 Solutions
1.1.1.2 Omitting details of the computation, with z = x + iy the answers are:
• <z4 = x4 − 6x2y2 + y4, =z4 = 4xy(x2 − y2).
• <z−1 = x/(x2 + y2), =z−1 = −y/(x2 + y2).
• <(z − 1)/(z + 1) = (x2 + y2 − 1)/((x + 1)2 + y2), =(z − 1)/(z + 1) = 2y/((x + 1)2 + y2).
• <z−2 = (x2 − y2)/(x2 + y2)2, =z−2 = −2xy/(x2 + y2)2.
1.1.1.3 This is an easy direct computation, or can be done by noting that all of these numbers have modulus 1, the firstpair have argument π ± π3 (so multiplication by 3 gives multiples of 2π), while the other pair in the second sethave arguments ± π3 (so multiplication by 6 gives multiples of 2π).
1.1.2.4 One may apply the quadratic formula.
1.1.4.3 If |a| = 1 then ∣∣∣∣ a − b1 − ab
∣∣∣∣ = |a|∣∣∣∣ a − ba − |a|2b
∣∣∣∣ = 1
and an analogous argument using b holds if |b| = 1. The only exception is when the denominator vanishes,which occurs iff ab = 1. In the case that either |a| = 1 or |b| = 1 we conclude immediately that this occurs iffa = b.
1.1.4.4 An equation in complex variables is two equations in real variables, in this case two simultaneous linear equa-tions. A fast way to extract them is to take the complex conjugate, so we have
az + bz + c = 0az + bz + c = 0.
Multiplying by a and b respectively,
|a|2z + abz + ac = 0
abz + |b|2z + bc = 0
the difference eliminates z and we obtain (|a|2 − |b|2
)z +(ac − bc
)= 0
so that there is a unique solution iff |a| , |b|, in which case the solution is
z =bc − ac|a|2 − |b|2
1.1.5.1 Each of the following statements is equivalent. ∣∣∣∣ a − b1 − ab
∣∣∣∣2 < 1
a − b1 − ab
a − b1 − ab
< 1
|a|2 + |b|2 − ba − ab < 1 − ab − ab + |a|2|b|2
|a|2(1 − |b|2
)< 1 − |b|2(
|a| < 1 and |b| < 1)
or(|a| > 1 and |b| > 1
).
1.1.5.3 By induction and the triangle inequality∣∣∣∣ n∑j=1
λ ja j
∣∣∣∣ ≤ n∑j=1
∣∣∣λ j
∣∣∣∣∣∣a j
∣∣∣ < n∑j=1
∣∣∣λ j
∣∣∣ = 1.
1.1.5.4 It is easiest to solve this problem if you recognize it as the equation of an ellipse with foci at ±a, as then youknow what to expect. In any case it is easy to see
2|a| = |a + a| = |a − z + a + z| ≤ |z − a| + |z + a| = 2|c|
so that |a| ≤ |c| is necessary for the existence of z satisfying the equation. Then for |c| ≥ |a|we may set z = |c|a/|a|and find that |z − a| = |c| − |a|, |z + a| = |c| + |a|, so |z − a| + |z + a| = 2|c|, whence it is also sufficient.
To find the maximum of |z| on the curve is easy, because repetition of our first argument shows |z| ≤ |c| on thecurve, but the point z = |c|a/|a| has |z| = |c|, so this maximum is achieved. The minimum is a little trickier. Oneway is as follows:
4|z|2 = 4zz =((z − a) + (z + a)
)((z − a) + (z + a)
)= |z − a|2 + |z + a|2 + 2|z|2 − 2|a|2
so 2(|z|2 − |a|2
)= |z − a|2 + |z + a|2. Then use the general inequality 2xy ≤ x2 + y2 for x, y ∈ R to see
2|z − a||z + a| ≤ |z − a|2 + |z + a|2 with equality iff |z − a| = |z + a|, and substitute into the square of the equationfor 2|c| to obtain 2|c|2 ≤ |z − a|2 + |z + a|2. Combining these we have 2|c|2 ≤ 2
(|z|2 − |a|2
), so |z|2 ≥ |c|2 − |a|2,
with equality iff |z − a| = |z + a|. The latter fact allows us to guess that a point at distance√|c|2 − |a|2 from 0 in
a direction perpendicular to a will lie on the curve, a guess that is rapidly verified using Pythagoras’ theorem.
1.2.1.1 The easy way to do this is to note that symmetry is preserved under rotation by π/4, such that the lines becomethe axes, then use that the symmetric points with respect to the axes are obtained by ± and conjugation, androtate back. The points are then
e−iπ/4(±eiπ/4a) = ±a,
e−iπ/4(±eiπ/4a) = ±e−iπ/4e−iπ/4a = ∓ia.
1.2.1.2 The vertices of an equilateral triangle are mapped to the vertices of an equilateral triangle by any map of theform z 7→ α(z + β), and for any given equilateral triangle there is a map of this form that takes it to the trianglewith vertices 1, e2πi/3, e−2πi/3. The correct choice is 3β = a1 + a2 + a3 and α = 3/(2a1 − a2 − a3). It is obviousthat the equality
a21 + a2
2 + a23 = a1a2 + a2a3 + a3a1
is preserved by the map z 7→ αz, and readily verified that it is preserved by z 7→ z + β. Thus it suffices to verifythe equality for the triangle with vertices 1, e2πi/3, e−2πi/3, in which case both sides are easily checked to sum to2.
1.2.2.2 If z = eiφ then
1 + cos φ + . . . + cos nφ = <(1 + z + . . . + zn)
= <1 − zn+1
1 − z
= <1 − z − zn+1 + zn
1 + |z|2 − z − z
=1 − cos φ − cos(n + 1)φ + cos nφ
2 − 2 cos φ
sin φ + . . . + sin nφ = =1 − z − zn+1 + zn
1 + |z|2 − z − z
=sin φ − sin(n + 1)φ + sin nφ
2 − 2 cos φ
Page 2
1.2.2.4 Let w = cos(2πi/n) + i sin(2πi/n) = e2πi/n. Provided h is not a multiple of n we have wnh , 1 and thus
1 + wh + . . . + w(n−1)h =1 − wnh
1 − wh =1 − e2πhi
1 − wh = 0.
1.2.3.1 To see when az + bz + c = 0 represents a line, recall exercise 1.1.4.4. We must have infinitely many solutionsfor the simultaneous equations, so they must be the same, so |a| = |b| and ac = bc. An equivalent form of thelatter is that c is a real multiple of (a + b); we may find this by summing it with its complex conjugate to find(a + b)c = (a + b)c, from which the ratio (a + b)/c is real. To be certain that the solution set is one dimensionalnot two dimensional we must insist that |a| , 0.
1.2.3.5 Rotation around the origin preserves the circle |z| = 1, preserves angles, and may be used to move a to lie onthe positive real axis. WLOG a > 1. Then the circle through a and 1
a has center 12(a + 1
a)
and radius 12(a − 1
a).
The circles intersect at right angles iff the radii to the intersection point are at right angles iff (by Pythagoras’theorem) the sum of the squares of the radii of the circles is the square of the distance to the center of the thecircle through a and 1
a . Written algebraically, this is equivalent to
1 +14(a −
1a)2= 1 +
14(a2 +
1a2 − 2
)=
14(a2 +
1a2 + 2
)=
14(a +
1a)2
1.2.4.5 I think it likely that this problem will have caused difficulties. The reason is that the spherical distance inequation (28) on page 20 of the book is not a geodesic metric! It is the metric corresponding to chords of thesphere, so the triangle inequality is always strict for three distinct points because they cannot all lie on a chord(the chord intersects the sphere at two points only). This and the fact that the center of the disc in the plane isnot the same as the center of the circle on the Riemann sphere mean that you cannot do either of the following:(1) take the distance d(a, a + Ra/|a|), or d(a, a − Ra/|a|) (because a is not the center in the spherical metric), (2)take half of d(a−Ra/|a|, a+Ra/|a|) (because the metric is not geodesic, so half of the diameter is not the radius).You can do one thing to simplify your calculations, which is to rotate the plane (an isometry of the sphericalmetric) so a ∈ R. At this point you have the option of finding the images on the sphere (using equations (25)and (26) on page 18), then find the center, then find the length of the chord corresponding to the radius. Thislatter can also be done by Pythagoras’ theorem, which makes the problem much simpler.
Suppose that the length of the chord joining the images of a ± R on the sphere is 2s. Then the distance fromthe center of the sphere to the midpoint of the chord is
√1 − s2 (by Pythagoras), so the distance radially from
this midpoint to the sphere is 1 −√
1 − s2. The landing point of the ray is the spherical center of the disc, andthe spherical radius we seek is the hypotenuse of the right triangle with vertices the landing point, the midpointof the chord, and the image of a + R. Since we have just seen the sides of this triangle have lengths s and1 −√
1 − s2 the hypotenuse has length 2(1 −√
1 − s2) (Pythagoras again). So it suffices to find s. By formula28 on page 20,
2s =2|(a + R) − (a − R)|(
1 + (a + R)2)(1 + (a − R)2) = 4R(1 + (a + R)2)(1 + (a − R)2) .
We can then compute
1 − s2 =
(1 + (a + R)2)(1 + (a − R)2) − 4R2(
1 + (a + R)2)(1 + (a − R)2)=
1 + (a + R)2 + (a − R)2 + (a2 − R2)2 − 4R2(1 + (a + R)2)(1 + (a − R)2)
=1 + 2(a2 − R2) + (a2 − R2)2(1 + (a + R)2)(1 + (a − R)2)
=
(1 + a2 − R2)2(
1 + (a + R)2)(1 + (a − R)2)
Page 3
so that
Spherical radius of disc = 2(1 −√
1 − s2)
= 2(1 −
∣∣∣1 + a2 − R2∣∣∣(
1 + (a + R)2)1/2(1 + (a − R)2)1/2 )= 2( √(1 + a2 − R2)2 + 4R2 − |1 + a2 − R2|√
(1 + a2 − R2)2 + 4R2
).
Page 4
Math 5120: Complex analysis. Homework 2 Solutions2.1.2.4 Suppose f = u+ iv is analytic and | f | is constant, so | f |2 = u2+ v2 = c. Analyticity
gives existence of the partial derivatives ux, uy, vx, vy and validity of the Cauchy-Riemann equations. Differentiating | f |2 = c gives uux + vvx = uuy + vvy = 0;Cauchy-Riemann lets us rewrite these as(
u vv −u
) (ux
vx
)=
(00
).
Since the determinant is | f |2 = c, we discover that either c = 0, whence f = 0everywhere, or the matrix is invertible and ux = vx = 0, from which also uy = vy =
0 by Cauchy-Riemann and thus u and v, and hence f , are constant.2.1.2.5 Each of the following statements is equivalent.
f (z) is analytic at z = z0.
limz→0
f (z + z0) − f (z0)z
exists.
limz→0
f (z + z0) − f (z0)z
exists.
limw→0
f (w + w0) − f (w0)w
exists. (Set w = z, w0 = z0)
f (w) is analytic at w = w0.
2.1.4.1(a) Let R(z) = z4(z3 − 1)−1. It has poles at ∞, 1, e2πi/3, e2πi/3. Set τ = 2πi/3. We needto expand at each pole, for which purpose (using the notation in the book) performeach of the following expansions. We include only the relevant terms, noting thepresence of those with lower order using an ellipsis.
R(z) = z +z
z3 − 1so G(z) = z
R(1 + z−1) =(1 + z−1)4
(1 + z−1)3 − 1=
(z + 1)4
z(z + 1)3 − z4 =z4 + · · ·
3z3 + · · ·=
z3+ · · ·
G1
( 1z − 1
)=
13(z − 1)
R(eτ + z−1) =(eτ + z−1)4
(eτ + z−1)3 − 1=
(eτz + 1)4
z(eτz + 1)3 − z4 =e4τz4 + · · ·
3e2τz3 + · · ·=
e2τz3+ · · ·
Geτ( 1z − eτ
)=
e2τ
3(z − eτ)
R(e2τ + z−1) =(e2τ + z−1)4
(e2τ + z−1)3 − 1=
(e2τz + 1)4
z(e2τz + 1)3 − z4 =e8τz4 + · · ·
3e4τz3 + · · ·=
e4τz3+ · · · =
eτz3+ · · ·
Ge2τ
( 1z − e2τ
)=
eτ
3(z − e2τ)
Thus
R(z) = z +1
3(z − 1)+
e4πi/3
3(z − e2πi/3)+
e2πi/3
3(z − e4πi/3).
1
2
2.1.4.1(b) In a similar but more tedious calculation, R(z) = z−1(z + 1)−2(z + 2)−3 has poles at0, −1 and −2. We perform the following expansions, including only the relevantterms.
R(z−1) =1
z−1(z−1 + 1)2(z−1 + 2)3 =z6
(1 + z)2(1 + 2z)3 =z6
8z5 + · · ·=
z8+ · · ·
G0
(1z
)=
18z
R(−1 + z−1) =1
(−1 + z−1)z−2(z−1 + 1)3 =z6
(z − 1)(1 + z)3 =z6
−z4 − 2z3 + · · ·= −z2 + 2z + · · ·
G−1
( 1(z + 1)
)=−1
(z + 1)2 +2
(z + 1)
R(−2 + z−1) =1
(−2 + z−1)(−1 + z−1)2z−3 =z6
(−2z + 1)(−z + 1)2
=z6
−2z3 + 5z2 − 4z + · · ·=−z3
2−
5z2
4−
17z8+ · · ·
G−2
( 1(z + 2)
)=
−12(z + 2)3 −
54(z + 2)2 −
178(z + 2)
Thus
R(z) =18z+−1
(z + 1)2 +2
(z + 1)−
12(z + 2)3 −
54(z + 2)2 −
178(z + 2)
.
2.1.4.4 Suppose R(z) is a rational function with |R(z)| = 1 on |z| = 1. Then
S (z) =1
R( 1
z)
is also a rational function, and since w = 1w whenever |W | = 1 we see that R(z) =
S (z) on |z| = 1. But then R(z) = S (z), because their difference is a rational functionwith zeros at every point of the unit circle. Now observe that if R(z) has a root ata of order α then S (z) has a pole at z = (a)−1 with the same order, so R(z) = S (z)has a pole of this order at this location. This argument is reversible; if R(z) hasa pole at b of order β then S (z) has a zero at z = (b)−1 with the same order, soR(z) = S (z) has a zero of this order at this location. Thus the poles and zeros ofR are bijectively paired by the map z 7→ (z)−1. (Note that this implies there areneither poles nor zeros at points of |z| = 1, because for these points z = (z)−1 andthere cannot simultaneously be a zero and a pole at a single point.) An equivalentformulation is that R(z) is necessarily a product of factors of the form z−a
1−az with|a| , 1. Since we proved in Exercise 1.1.4.3 that such factors have modulus 1on the unit circle, this condition is also sufficient. Grouping the factors and usingnegative powers α j when a zero is outside the unit disc we see that the generalform of the rational function we seek is
R(z) = czkm∏
j=1
( z − a j
1 − a jz
)α j
with k and α j in Z and |a j| < 1 for all j.
3
2.1.4.6 Suppose R(z) = cP(z)/Q(z) is a rational function, where P and Q are monic poly-nomials without common zeros. Let m = deg(P) and n = deg(Q). It will beconvenient at first to assume neither P nor Q is constant. We have
R′(z) = cP′(z)Q(z) − P(z)Q′(z)
Q2(z).
It is an easy observation that deg(P′) = m−1 and deg(Q′) = n−1, thus deg(P′Q) =deg(PQ′) ≤ m + n − 1. Observe that the lead term in P′Q has coefficient n whilethat in PQ′ has coefficient m. If we assume m , n then these terms cannot cancel,so deg(P′Q − PQ′) ≤ m + n − 1 with equality at least when m , n. To find thedegree of R′(z) we must then cancel common factors. Any root of Q with order αis a root of P′Q with the same order, so to be a root of P′Q − PQ′ it must divideP or Q′. By assumption P and Q have no common roots, so it must be a root ofQ′. An easy computation shows that Q and Q′ have common roots if and only ifQ has multiple roots, and that the order of the root in Q′ is α−1. We conclude thatQ′/Q is of the form S/T , where T is the (monic) product of the distinct factors ofQ and has order k ≤ n, and deg S = k − 1. It is then apparent that
R′(z) = cP′(z)T (z) − P(z)S (z)
Q(z)T (z)with deg(P′T − PS ) ≤ m + k − 1 (with equality if m , n), deg(QT ) = n + k, andthe numerator and denominator having no common roots. It follows that
deg(R′) ≤ maxm+k−1, n+k = maxm−1, n+k ≤ maxm−1, n+n ≤ deg R+n ≤ 2 deg(R)
and that equality holds at the first inequality if m , n, at the second if k = n (i.e.all poles of R are distinct), at the third and fourth if deg R = n ≥ m. Thus we havedeg(R′) ≤ 2 deg(R) and conditions under which equality holds.
We may also get lower bounds. If m , n then we have equality in the firstinequality above, so deg(R′) = maxm + k − 1, n + k ≥ max m, n + 1 ≥ deg(R)because we assumed Q non-constant and thus k ≥ 1. If m = n then deg(P′T −PS ) ≤ m + k − 1 < n + k = deg(QT ), so deg(R′) = n + k ≥ deg(R) + 1. (Note thisalso implies that deg(R′) = 2 deg(R) when m = n = k
What remains are the special cases where either P or Q is constant. If Q isconstant and P is not then R is a polynomial, so deg(R′) = deg(R) − 1. If P isconstant and Q is not then the above reasoning implies R′ = S/QT , so deg(R′) =deg(QT ) = k + n > n = deg(R). If both are constant then deg(R′) = −∞.
We may summarize our results as follows. If R(z) is a non-constant rationalfunction, then deg(R) − 1 ≤ deg(R) ≤ 2 deg(R), with equality on the left iff R is apolynomial and equality on the right at least when the poles of R are distinct andthere is no pole at infinity.
2.2.3.3 Suppose∑|a j| converges. A re-ordering of the sum is given by a bijection η : N→
N, where the new sum is∑
j aη( j). Both sums converge absolutely; set S =∑
j a j
and T =∑
j aη( j). Given ε > 0 let N be so large that for any K ≥ N, each of∑j>K |a j| ≤ ε,
∣∣∣S − ∑j≤K a j
∣∣∣ ≤ ε and∣∣∣T − ∑
j≤K aη( j)∣∣∣ ≤ ε. Let M be so large that
η( j) : j ≤ M⊃ j ≤ N (Note that then M ≥ N.)
|S − T | ≤∣∣∣∣∑j≤M
a j −∑j≤M
aη( j)
∣∣∣∣ + 2ε ≤∑j>N
|a j| + 2ε ≤ 3ε
where middle inequality used that the sum is over those j in( j ≤ M \ η( j) : j ≤
M)∪
(η( j) : j ≤ M \ j ≤ M
), which does not include any j ≤ N.
4
2.2.3.4 For fixed z with |z| ≥ 1 we have |nzn| = n|z|n → ∞, while if |z| < 1 we haven|z|n = exp(log n + n log |z|)→ 0 because log |z| < 0 and (n log |z|)/ log n→ −∞ asn → ∞ (using, for example, L’Hopital’s rule). Hence nzn converges pointwiseon |z| < 1. A slight improvement is that on |z| ≤ r < 1 we have n|z|n ≤ nrn → 0, sothat the convergence is uniform on this disc. Any compact set that lies in |z| < 1 isin such a disc, so we have nzn converges uniformly to the zero function on anycompact subset of the open unit disc. Moreover, this is the largest collection ofsets on which the convergence is uniform. To see this, suppose S is a set on whichnzn converges uniformly, from which we see S is in the unit disc and n|z|n → 0.If S is not a compact subset of the disc then it contains points zk such that |zk | → 1.If n is so large that n|z|n < 1/2 on S then the fact that n|zk |
n → n as k → ∞ providesa contradiction.
2.2.3.6 There are a number of ways of doing this, the most usual being to use summationby parts. I will show a method that is messier, but perhaps easier to visualize. LetU =
∑j u j, V =
∑j v j; suppose WLOG that
∑j u j is the absolutely convergent
series and set W =∑
j |u j|. It is helpful to arrange the terms in a doubly-infinitelist.
(u jvk) j,k =
u1v1 u1v2 u1v3 . . .u2v1 u2v2 u2v3 . . .u3v1 u3v2 u3v3 . . ....
......
We see immediately that
∑n−1j=1 u jvn− j is the sum along the nth upward diagonal,
so∑m
n=1∑n−1
j=1 u jvn− j is the sum over the upper left triangle j + k ≤ m in the list.We can also see that UV may be approximated by the sum over a square j ≤ p,k ≤ p, because for fixed ε > 0 we can take N so p ≥ N implies
∣∣∣U −∑j≤p u j
∣∣∣ ≤ ε,∣∣∣V −∑k≤p vk
∣∣∣ ≤ ε, and so∣∣∣∣∣UV−(∑
j≤p
u j
)(∑k≤p
vk
)∣∣∣∣∣ = ∣∣∣∣∣UV−V∑j≤p
u j+V∑j≤p
u j−(∑
j≤p
u j
)(∑k≤p
vk
)∣∣∣∣∣ ≤ |V |ε+∣∣∣∣∣∑j≤p
u j
∣∣∣∣∣ε ≤ ε(|V |+∑j≤p
|u j|
)≤ ε
(|V |+W
).
Thus the difference between UV and the sum we are considering is controlled bya small term plus the sum over a region inside an upper left triangle and outside asquare, i.e. j + k ≤ m but also j ≥ p, k ≥ p.∣∣∣∣∣UV−
m∑n=1
n−1∑j=1
u jvn− j
∣∣∣∣∣ ≤ ε(|V |+W)+
∣∣∣∣∣ m∑n=1
n−1∑j=1
u jvn− j−(∑
j≤p
u j
)(∑k≤p
vk
)∣∣∣∣∣ ≤ ε(|V |+W)+
∣∣∣∣∣ ∑ j≥p,k≥p, j+k≤m
u jvk
∣∣∣∣∣.We split this sum into two pieces, one with k ≥ p and one with k ≤ p. We want abound independent of m, and you should think that m p. Since it is a finite sumit can be rearranged any way we like; we will sum first along the rows and thendown the columns. For the piece with k ≥ p the sum along the jth row is boundedas follows: ∣∣∣∣∣u j
m− j∑k=p+1
vk
∣∣∣∣∣ ≤ |u j|ε
provided p is so large that∣∣∣∑q
p+1 vk
∣∣∣ ≤ ε for all q ≥ p; this can be achieved by (ifnecessary) increasing N, because
∑vk is convergent. Summing over all relevant
5
rows we have a contribution bounded as follows:∣∣∣∣∣m−p∑j=1
u j
m− j∑k=p+1
vk
∣∣∣∣∣ ≤ ε m−p∑j=1
|u j| ≤ ε
∞∑j=1
|u j| = Wε.
Now for the piece with k < p we reason as follows. Since∑
k vk converges,∑
k≤r vk
is a convergent sequence in r, so has absolute value bounded by a constant X. Nowthe sum across each row in the second piece is of size∣∣∣∣∣u j
minp−1,m− j∑1
vk
∣∣∣∣∣ ≤ X|u j|
and summing over the rows we have∣∣∣∣∣ m∑j=p+1
u j
minp−1,m− j∑1
vk
∣∣∣∣∣ ≤ Xm∑
j=p+1
|u j| ≤ X∞∑
j=p+1
|u j| ≤ Xε
provided p is large enough that∑
j≥p+1 |u j| ≤ ε; again this can be achieved byincreasing N. Combining our estimates we now have∣∣∣∣∣UV −
m∑n=1
n−1∑j=1
u jvn− j
∣∣∣∣∣ ≤ ε(|V | + 2W + X)
and the result follows.By the way, if you are wondering how this is connected to the material in this
chapter, the following may be of interest. Consider the functions U(z) =∑
j u jz j
and V(z) =∑
j v jz j. These converge at z = 1, so have radius of convergence atleast 1; moreover Abel’s limit theorem says that U(r) → U(1) and V(r) → V(1)as r → 1, where r ∈ (0, 1). On the disc |z| < 1 they are absolutely convergent, andtheir product U(z)V(z) has a power series expansion
∑j cnzn. We can compute by
the Leibniz rule
cn =1n!
dn
dzn
(U(z)V(z)
)∣∣∣∣z=0=
1n!
n∑j=0
n!j!(n − j)!
d jU(z)dz j
∣∣∣∣z=0
dn− jV(z)dzn− j =
n−1∑j=1
u jvn− j.
If we now knew that∑
n cn =∑
n∑n−1
j=1 u jvn− j was convergent, then Abel’s limittheorem would imply
∑n cn = limr→1 U(r)V(r) = limr→1 U(r) limr→1 V(r) =
U(1)V(1), which is the theorem we seek. This suggests that there is a proof essen-tially by the proof of Abel’s limit theorem, which is the standard summation byparts proof.
2.2.4.2
2z + 3z + 1
=2(z − 1) + 5(z − 1) + 2
= 2 +12
11 + (z − 1)/2
= 2 +12
∞∑j=0
(−1) j
2 j (z − 1) j if |z − 1| < 2.
The radius of convergence is 2.
6
2.2.4.4 The following statements are equivalent.
∑n
anzn has radius of convergence R
lim sup |an|1/n = R−1
lim sup |an|1/2n = R−1/2 and lim sup |an|
2/n = R−2∑n
anz2n has radius of convergence√
R, and∑n
a2nzn has radius of convergence R2
2.2.4.8 The series∑∞
0 zn(1 + z)−n converges if and only if |z(1 + z)−1| < 1, which is if andonly if |z| < |1 + z|, if and only if<(z) > 1
2 .2.3.2.2
cos iz =e−z + ez
2= cosh z
sin iz =e−z − ez
2i= i sinh z
We can get addition formulae for cosh and sinh by computing
cosh a cosh b =ea + e−a
2eb + e−b
4=
ea+b + e−(a+b) + ea−b + eb−a
4
sinh a sinh b =ea − e−a
2eb − e−b
4=
ea+b + e−(a+b) − ea−b − eb−a
4
cosh a sinh b =ea + e−a
2eb − e−b
4=
ea+b − e−(a+b) − ea−b + eb−a
4
elementary algebra implies
cosh(a + b) = cosh a cosh b + sinh a sinh b cosh 2a = cosh2 a − sinh2 a
sinh(a + b) = sinh a cosh b + cosh a sinh b sinh 2a = 2 sinh a cosh a
and substitution of the above expressions for cos and sin retrieves the usual trigono-metric addition formulae, which could alternatively be used to obtain the above.
2.3.2.3
cos(x + iy) = cos x cos iy − sin x sin iy = cos x cosh y − i sin x sinh y
sin(x + iy) = sin x cos iy + cos x sin iy = sin x cosh y + i cos x sinh y
7
2.3.4.4
ez = 2 when z = log 2 + 2kπi, k ∈ Z
ez = −1 when z = (1 + 2k)πi, k ∈ Z
ez = i when z =(12+ 2k
)πi, k ∈ Z
ez =−i2
when z = − log 2 +(32+ 2k
)πi, k ∈ Z
ez = −1 − i when z =12
log 2 +(54+ 2k
)πi, k ∈ Z
ez = 1 + 2i when z =12
log 5 +(arctan 2 + 2k
)πi, k ∈ Z and arctan in (0, π/2).
2.3.4.5 Let z = x + iy, so ez = ex cos y + iex sin y. Then the real and imaginary parts ofexp(ez) are obtained from
exp(ez) = exp(ex cos y + iex sin y) = exp(ex cos y) cos(ex sin y) + i exp(ex cos y) sin(ex sin y)
2.3.4.6
2i = exp(i log 2) = cos(log 2) + i sin(log 2) single valued because 2 ∈ R
ii = exp(i log i) = exp(−π
2− 2kπ), k ∈ Z
(−1)2i = exp(2i log(−1)) = exp(−2π − 4πk), k ∈ Z.
Note that the last example shows something we have lost in making the conventionthat the logarithm is single valued on the positive reals and multivalued elsewhere,because (−1)2i , ((−1)2)i = 1, but simply contains 1 as one of its values.
Math 5120: Complex analysis. Homework 3 Solutions
3.1.2.6 R and C are separable metric spaces, so see Exercise 3.1.3.7 below.
3.1.3.4 Let A = (0, y) : |y| ≤ 1 and B = (x, sin(1/x) : x > 0 in R2. Note that both A and B are continuous images ofconnected subsets of R, so are connected. If there is a disconnection of A ∪ B, then the set containing a pointof A must contain all of A because A is connected, and similarly for B. So the disconnection must consist ofopen sets U ⊃ A and V ⊃ B. But then U contains a disc around 0, and any such disc contains points from B, forexample those of the form
((2πn)−1, sin(2πn)
)for all sufficiently large n ∈ N, so U ∩ V , ∅ and there is no such
disconnection. We conclude that A ∪ B is connected. (Note: A ∪ B is not path connected).
3.1.3.5 Let En = (x, 1/n) : 0 ≤ x ≤ 1 and E∞ = (x, 0) : 0 ≤ x ≤ 1 in R2. Let E = E∞ ∪(∪nEn
). For any n let Un
be the open set consisting of all points within distance (n + 1)−2 of En, and Vn be the interior of the complementof Un. Then Un ∩ Vn = ∅, Un ⊃ En and Vn ⊃ E \ En. We conclude that the maximal connected set containing apoint of En is contained in En. Since En is connected (because it is a curve), we conclude that En is a componentof E for each n. For a point in E∞ we see that the component containing this point must contain E∞, becauseE∞ is connected; from the above it does not contain any point of En for any n, so E∞ is a component. We havetherefore found the decomposition of E into its components. We have also seen that the components of the formEn are closed in R2 and relatively open in E; the component E∞ is closed in R2, but not relatively open in Ebecause any open neighborhood of E∞ in R2 intersects some En.
Finally, a set is locally connected if every neighborhood of a point in the set contains a connected neighborhoodof the point. Consider (0, 0) ∈ E∞. Any neighborhood of this point is the intersection of E with an open set inR2, and therefore cannot be connected because it contains points from another component En. Since this pointhas no connected neighborhoods it is not locally connected. (Remark: local connectivity of a set is equivalentto all components being open.)
3.1.3.7 S is discrete in a metric space (M, d), so for all x ∈ S there is rx such that the ball around x of radius rx, denotedB(x, rx), has B(x, rx)∩ S = x. Observe that then B(x, rx/2)∩ B(y, ry/2) = ∅ if x , y, because if there is z in theintersection we would have
|x − y| ≤ |x − z| + |y − z| < (rx + ry)/2 ≤ maxrx, ry
contradicting either x 3 B(y, ry) or y 3 B(x, rx).
Now if A ⊂ M is a dense set then for each x there is ax ∈ B(x, rx/2) ∩ A, and ax , ay if x , y becauseB(x, rx/2) ∩ B(y, ry/2) = ∅. The map x 7→ ax is then an injection from S to A, so the cardinality of S cannotexceed that of A. In particular if M is separable we may take A to be a countable dense subset and conclude Sis countable.
3.2.2.1 We want to give a domain and a definition of√· so that
√1 + z +
√1 − z is a single-valued analytic function.
In the book there is a definition of a single valued analytic branch of√
w for w ∈ C \ (−∞, 0]. It suffices thenthat we ensure 1 + z and 1 − z are in this set, which is true if z ∈ C \
((−∞,−1] ∪ [1,∞)
). Nothing more need be
done, as on this set√
1 + z +√
1 − z is a sum of single-valued analytic functions.
Math 5120: Complex analysis. Homework 4 Solutions3.3.1.1 The map z 7→ z fixes 0,1 and ∞. Consider a fractional linear transformation z 7→
az+bcz+d fixing these points. Fixing 0 implies b = 0, fixing ∞ implies c = 0, fixing 1implies a/d = 1, so the map is the identity. Thus z 7→ z is not a fractional lineartransformation.
3.3.3.1 A reflection z 7→ z∗ = g(z) is defined via (z∗, z1, z2, z3) = (z, z1, z2, z3). Let φ bethe fractional linear transformation so φ(z1) = 1, φ(z2) = 0 and φ(z3) = ∞, and letf (z) = z. Then φ(z∗) = φ(z), so g = φ−1 f φ. Now we know from the bookthat φ maps circles to circles (on the Riemann sphere), so it suffices to see that thesame is true of f . A circle not through∞ has the form |z − a| = r, so is mapped tor = |z − a| = |z − a|, which is also a circle. A circle through ∞ is z = x + iy withax+by+ c = 0, a, b, c ∈ R, and is mapped to z = x+ iy with ax−by+ c = 0, whichis also a circle through∞.
3.3.3.5 It is easy to think of some fractional linear transformations that fix |z| = R as a set.For example, any rotation around the origin (i.e. z 7→ az with |a| = 1), the mapz 7→ R2/z. In the case R = 1 we also know from prior homeworks that z 7→ z−a
1−az
for |a| , 1 preserves |z| = 1, so z 7→ R (z/R)−a1−a(z/R) =
Rz−aR2
R−az preserves |z| = R. Themore difficult question is whether we have found all of the maps or not. One wayto approach the problem is to take an arbitrary fractional linear transformation thatfixes |z| = R, compose it with some of the maps we know, and try to get the resultto fix 3 points, so that the composition is the identity.
Suppose that φ is a fractional linear transformation that fixes |z| = R as a set.Let eiθ = φ(R)/R, so z 7→ e−iθφ(z) fixes the point z = R. Let a = e−iθφ(0)/R andψ(z) = Rz−aR2
R−az . Then z 7→ ψ(e−iθφ(z)) fixes 0 and R, and also fixes |z| = R as aset. Since the mapping of reflection in a circle depends only on the circle, thismap must preserve pairs of points that are symmetric under reflection in |z| = R.Now 0,∞ has such reflective symmetry and ψ(e−iθφ(0)) = 0, so we concludeψ(e−iθφ(∞)) = ∞. Thus we have a fractional linear transformation fixing threepoints, which we know is the identity, and so ψ(e−iθφ(z)) = z, from which φ(z) =eiθψ−1(z). It is easy to compute that ψ−1(z) = Rz+aR2
R+az . We have thus shown that
φ(z) = eiθ Rz + aR2
R + az.
3.3.3.7 If we had a fractional linear transformation taking |z| = 1 and |z − 14 | =
14 to
concentric circles, then by composing with a translation to move their commoncenter to 0 and a dilation we could assume that the fractional linear transformationpreserves the unit circle; note that neither operation affects the ratio of radii ofthe image circles. By exercise 3.3.3.5 with R = 1, such a map is of the formz 7→ eiθ z−a
1−az = f (z), |a| , 1. The rotation preserves both the concentric propertyand the ratio of radii, so we are free to choose it, for example by setting θ = 0.Now observe that the initial and final configurations are mapped to themselves byz 7→ z, so g(z) = f (z) = e−iθ z−a
1−az also preserves |z| = 1 and takes |z − 14 | =
14 to a
circle with center at 0. If the image of |z− 14 | =
14 under f is |z| = r, it follows from
exercise 3.3.3.5 that
f g−1(z) =z+a1+az − a
1 − a z+a1+az
=(1 − a2)z + (a − a)(1 − |a|2) + (a − a)z
=1 − a2
1 − |a|2z + a−a
1−a2
1 + a−a1−|a|2 z
1
2
is a fractional linear transformation fixing |z| = r, so is of the form z 7→ eiφ rz+r2br+bz .
We conclude that a ∈ R. Knowing this, we see that f preserves R, and thereforethat f (0) and f ( 1
2 ) are the points ±r. This gives us that simultaneously −a = ±rand ( 1
2 − a)/(1 − a2 ) = ∓r, which we reduce to r2 ± 4r + 1 = 0, or (finding only
solutions with r > 0), a = r = 2 ±√
3. We have found necessary conditions forf to map |z − 1
4 | =14 to |z| = r, and it remains to see they are sufficient. But f
is a fractional linear transformation so maps circles to circles; by construction theimage of |z − 1
4 | =14 passes through ±r, and since a ∈ R the image is mapped
to itself by z 7→ z, so it is |z| = r. We have therefore determined that the map isone of z 7→ eiθ z−a
1−az for a = 2 ±√
3 and the ratio of the smaller to larger radius is2 −√
3 : 1 = 1 : 2 +√
3.3.3.3.8 We are now asked to repeat this problem for |z| = 1 and x = 2. All of the same
arguments apply to say that the map is z 7→ eiθ z−a1−az for a ∈ R. The points z = 2
and z = ∞ are invariant under z 7→ z, so go to ±r. Computing as in the previousproblem we find again r = 2 ±
√3, but with a = 1
r = 2 ∓√
3. Alternativelywe could notice that the map z 7→ 1
z takes the circle in 3.3.3.7 to the line in thisproblem (to see this, note what happens to 0 and ∞ and that the symmetry underz 7→ z is preserved), so the whole problem is obtained from 3.3.3.7 by compositionwith z 7→ 1
z .3.4.2.2 The circles intersect at z = 1 and are parallel there, so any fractional linear trans-
formation mapping 1 to∞will give us the region between two circles that intersectat ∞ with zero angle, meaning a strip between two parallel lines. There are manyfractional linear transformations taking 1 to∞, but it is convenient to take one thatmaps one of the circles to the real axis; for example z 7→ i z+1
z−1 takes |z| = 1 to Rbecause it takes −1 to 0 and i to 1. The image of the other line must have the formz = x + ic for a constant c; by computing 0 7→ −i we find c = −i. Now we canmultiply by −π and exponentiate to get the upper half plane. Our final map is
z 7→ exp(−πi
z + 1z − 1
).
3.4.2.3 The arc |z| = 1, y ≥ 0 can be mapped to a segment on the real line by a fractionallinear transformation, and we know how to map the complement of a segment onthe complement of the unit disc (or rather we know how to map the complement ofthe unit disc on the complement of the segment [−2, 2] using the map z 7→ z+ z−1).For the first step we should take a point on |z| = 1 to ∞, and since it is convenientto make the image segment symmetric around 0 we may as well take i to 0 and −ito∞. After multiplying by 2 so our arc goes to [−2, 2], the map is z 7→ 2i z−i
z+i .Now the trickier issue is how to invert z 7→ z+z−1. Formally using the quadratic
equation the inverse is z 7→ 12 (z ±
√z2 − 4), but we have to make sense of the
square root. If z2 − 4 mapped the complement of [−2, 2] to the complement of aray connecting 0 and∞ we could use a version of the usual square root, but it doesnot – the image in this case is the complement of the interval [−4, 0]. However,we can do whatever algebraic manipulation we desire to the expression for theformal inverse and still have a formal inverse. Knowing that we want the bit insidethe square root to omit a ray from 0 to ∞ (preferably the negative real axis) andthat right now what is in there omits [−4, 0], tends to suggest we should divide byeither z or z− 2 to move an endpoint to∞. A little playing with the formula yields
3
a different formal expression for the inverse branches
f±(z) =z2(1 ±√
1 − 4/z2)in which 1 − 4/z2 is easily seen to omit the negative real axis (and the ray (2,∞)on the positive real axis). We can therefore use the usual definition of
√· taking
C \ (−∞, 0] to the right half plane. The result is that the branches f±(z) are eachwell defined on the complement of [−2, 2].
Now note that the two branches of the inverse map either to the interior or theexterior of the unit disc, and that we want the one which maps to the exterior. Wecan verify that the desired one is z
2 (z +√
1 − 4z−2) by observing that it has imageconverging to∞ as z→ ∞.
Our final result is therefore that
z 7→12
(2i
z − iz + i
)(1 +(1 −
4(z + i)2
−4(z − i)2
)1/2)=
1 + izz + i
(1 +(1 +
(z + i)2
(z − i)2
)1/2)where we readily determine that for z not on our arc, 1+ (z+i)2
(z−i)2 is not on the negativereal axis.
3.4.2.7 We wish to map the exterior of an ellipse on the interior of the unit disc withpreservation of symmetries, those being z 7→ z and z 7→ −z. Recall that in studyingthe map z 7→ z + z−1 = f (z) we saw that the circle z = reiθ, is mapped to z = x + iywith x = (r + r−1) cos θ and y = (r − r−1) sin θ, which is an ellipse centered at0 with semi-major axis along the real line and of length a = r + r−1 and semi-minor axis along the imaginary axis and of length b = |r − r−1|; this is the ellipse(x/a)2 + (y/b)2 = 1, a > b. The map also preserves the symmetries of the ellipse,in that f (z) = f (z) and f (−z) = − f (z).
The map z 7→ f (z) is a double covering; if r > 1 then both |z| > r and |z| < 1/rare mapped onto the the exterior of the ellipse. It is easy to see that when r > 1we should have r = (a − b)/2 and thus 1/r = 2/(a − b). The inverse map has twobranches z 7→ z
2 (1 ±√
1 − 4z−2) which were discussed in the previous problem.Here we need the branch z 7→ z
2 (1 −√
1 − 4z−2) which maps the exterior of theellipse to the interior of the disc of radius 2/(a − b). There are only a few moredifficulties. The first is that we need the ellipse to have a > b, so that a preliminaryrotation is needed if b > a. The second is that we need r = (a − b)/2 > 1 inthe above analysis. To ensure that this is the case it is convenient to perform apreliminary dilation, for example one ensuring that r = 2. Then the inverse branchof f will map to the disc |z| < 1/2, and an additional dilation will bring us to|z| < 1. At this point it is simplest to split into cases.
Case 1: a > b. Then (x/a)2 + (y/b)2 = 1 has its longer axis along the real line.We perform the dilation z 7→ 4z/(a − b), so that the length of the semi-major axisbecomes 4a/(a−b) and of the semi-minor axis becomes 4b/(a−b); the exterior ofthis is the image of |z| < 1/2 under f (z), so applying z 7→ z
2 (1 −√
1 − 4z−2) takesit to |z| < 1/2 and dilating by a factor of 2 gives |z| < 1. The result is
z 7→ 24z
2(a − b)
(1 −(1 −
4(a − b)2
16z2
)1/2)=
4za − b
(1 −(1 −
(a − b)2
4z2
)1/2)Case 2: b > a. We can use the same argument as above, but must first rotate by
π/2 to get the semi-major axis along the real line, and must undo this at the end.The maps are z 7→ iz 7→ 4iz/(b−a), then the inverse branch, then dilation by 2 and
4
rotation by −i, resulting in
z 7→ −2i4iz
2(b − a)
(1 −(1 −
4(b − a)2
−16z2
)1/2)=
4zb − a
(1 −(1 +
(a − b)2
4z2
)1/2)Case3: b = a. Here we have just the circle |z| = a = b, so we can scale
by z 7→ z/a to the unit circle and invert to map the exterior to the interior. Theresulting map is
z 7→az.
Math 5120: Complex analysis. Homework 5 Solutions4.1.3.2 Using the parametrization, z = reiθ, so z = x + iy with x = r cos θ, y = r sin θ we
have ∫|z|=r
x dz =∫ 2π
0r cos θireiθ dθ
= ir2∫ 2π
0cos2 θ + i cos θ sin θ dθ
= ir2∫ 2π
0
12
(cos 2θ + 1) dθ − r2∫ 2π
0
12
sin 2θ dθ
= πir2.
Using x = 12 (z + r2z−1) as suggested in the text we use that
∫|z|=r z dz = 0 by
equation (11) on pg 107 and∫|z|=r z−1 dz = 2πi by the following discussion to
conclude∫|z|=r x dz = πir2.
4.1.3.3 The function (z2 − 1)−1 does not have an antiderivative on the disc |z| ≤ 2, so wecannot directly apply the results from section 3.3. Since we do not yet have theCauchy theorem, we must integrate directly using a parameter. This is messy, butthe point is to get you to appreciate the Cauchy theorem and residue calculus oncewe have them. Take z = 2eiθ, so the integral is∫ π
−π
2ieiθ
4e2iθ − 1dθ =
∫ π−π
2ieiθ(4e−2iθ − 1)(4e2iθ − 1)(4e−2iθ − 1)
dθ =∫ π−π
8ie−iθ − 2ieiθ
17 − 8 cos 2θdθ =
∫ π−π
6i cos θ + 10 sin θ17 − 8 cos 2θ
dθ.
The real part of the integral has an antisymmetric integrand on a symmetric intervalso is zero. For the imaginary part we note from symmetry and periodicity of cosinethat∫ π
−π
cos θ17 − 8 cos 2θ
dθ = 2∫ π
0
cos θ17 − 8 cos 2θ
dθ
= 2∫ π/2
0
cos θ17 − 8 cos 2θ
dθ + 2∫ π
pi/2
cos θ17 − 8 cos 2θ
dθ
= 2∫ π/2
0
cos θ17 − 8 cos 2θ
dθ + 2∫ 0
pi/2
cos(π − θ)17 − 8 cos(2π − 2θ)
(−dθ)
= 2∫ π/2
0
cos θ17 − 8 cos 2θ
dθ + 2∫ π/2
0
− cos(θ)17 − 8 cos(2θ)
dθ
= 0
so that the whole integral is zero.4.1.3.6 We know that f ′/ f should have antiderivative log f , provided that log f is well
defined. Suppose then that | f (z) − 1| < 1 for z ∈ Ω. We know that log is a single-valued analytic function on C \ (−∞, 0], with derivative z−1. Since the values off avoid (−∞, 0] it follows that log f is well defined and analytic, with derivative(by the chain rule) f ′(z)/ f (z). Hence the integrand is the derivative of an analyticfunction on Ω and by the result on page 107 the integral around any closed curvein Ω is zero.
4.1.3.8 In order that∫γ
log z = 0 is meaningful and true, we must first require that γlies in a region where log z is a single-valued function (otherwise the integral isnot well-defined). We know Ω = C \ (−∞, 0] is such a region (in fact, so is the
1
2
complement of any connected set containing 0 and∞). Now we might hope to findthat∫γ
log z = 0 for any closed curve in Ω by proving that log z is the derivative ofan analytic function. Since we know that (log z)′ = z−1 on Ω, it is easy to checkthat (z log z − z)′ = log z + z/z − 1 = log z there. Hence a suitable condition is thatγ is a closed curve in Ω.
4.2.2.1 Using Cauchy’s integral formula we have∫|z|=1
ez
zdz = 2πie0 = 2πi.
4.2.2.2 Writing 1z2+1 =
1(z+i)(z−i) =
12i
(1
z−i −1
z+i
)we can apply the Cauchy integral formula
to each of the below integrals∫|z|=2
1z2 + 1
dz =12i
∫|z|=2
1z − i
dz −12i
∫|z|=2
1z + i
dz = π − π = 0.
Compare this to exercise 4.1.3.3 above!
Math 5120: Complex analysis. Homework 6 Solutions4.2.3.1(a) By the Cauchy formula for derivatives,
∫|z|=1 z−n f (z) dz = 2πi f (n−1)(z)/(n−1)! pro-
vided f is analytic in the unit disc. Applying this to f (z) = ez gives∫|z|=1 z−nez dz =
2πi/(n − 1)!.4.2.3.1(b) If n and m are both non-negative then the integrand in
∫|z|=2 zn(1 − z)m dz is a
polynomial, hence analytic, and so the integral is zero by the Cauchy formula.If n is negative and m non-negative then our previous reasoning says the resultis 2πi f (|n|−1)(0)/(|n| − 1)! for f (z) = (1 − z)m, which is 0 if m < |n| − 1 and2πi(−1)|n|−1
(m|n|−1
)otherwise. Similarly, if n is non-negative and m negative then
we get 0 if n < |m| − 1 and 2πig(|m|−1)(1)/(|m| − 1)! = 2πi(
n|m|−1
)otherwise (with
g(z) = zn). If both m and n are negative it is a little more work, basically be-cause we do not have the Cauchy formula for general curves, so cannot reduceto a circle around 0 and another around 1. However there is a still a relativelyquick argument of the same type as above. We may use partial fractions to writezn(1 − z)m = P(z)zn + Q(z)(1 − z)m, where P has degree at most |n| − 1 and Qhas degree at most |m| − 1; note that P(z)(1 − z)−m + Q(z)z−n = 1. Integrating theP(z)zn term gives 2πiP(|n|−1)(0)/(|n| − 1)!, which is just 2πi times the leading coef-ficient p of P(z). Similarly integrating the Q(z)(1 − z)m term gives (−1)m2πi timesthe leading coefficient q of Q. These leading terms give the power of z−m−n−1 inP(z)(1− z)−m +Q(z)z−n to be (−1)m(p+ q), and this must be zero because m, n < 0implies −m − n ≥ 2, so −m − n − 1 ≥ 1. We may summarize our results as
∫|z|=2
zn(1 − z)m dz =
2πi(−1)−n−1
(m−n−1
)if n < 0 and m ≥ −n − 1
2πi(
n−m−1
)if m < 0 and n ≥ −m − 1
0 otherwise
4.2.3.1(c) We write |z − a|2 = (z − a)(z − a) = (z − a)((ρ2/z) − a
)on |z| = ρ and use the fact
that on z = ρeiθ we have |dz| = ρdθ and dz = ireiθdθ = izdθ, so |dz| = ρdz/iz. Thusthe integral is∫
|z|=ρ|z − a|−4 |dz| =
∫|z|=ρ
1(z − a)2
(ρ2
z− a)−2 ρ
izdz =
ρ
i
∫|z|=ρ
z(z − a)2(ρ2 − az)2 dz.
Now by assumption, |a| , ρ. The integrand has singularities at z = a and z = ρ2/a,which are reflection-symmetric in the circle |z| = ρ, so only one is inside the circle.The Cauchy formula then says that the integral is 2πρ f ′(a) with f (z) = z/(ρ2− az)2
if |a| < ρ, which evaluates to 2πρ(ρ2 + |a|2)/(ρ2 − |a|2)3. Similarly if |a| > ρit becomes 2πρg′(ρ2/a) with g(z) = z/(a)2(z − a)2, which is the negative of theprevious answer. Summarizing∫
|z|=ρ|z − a|−4 |dz| =
2πρ(ρ2+|a|2)(ρ2−|a|2)3 if |a| < ρ
−2πρ(ρ2+|a|2)
(ρ2−|a|2)3 if |a| > ρ.
4.2.3.2 We assume f is globally analytic and there is C so | f (z)| ≤ C|z|n for all sufficientlylarge |z|. Let P(z) be the (n−1)th order Taylor polynomial of f (z) at 0, so f (z)−P(z)has a zero of order n at 0 and therefore f (z) − P(z) = zng(z) for some globallyanalytic g(z). Since P is an order n − 1 polynomial we have |P(z)|/|z|n ≤ 1 for allsufficiently large |z|; using the hypothesis we get
|g(z)| ≤| f (z)||z|n+|P(z)||z|n
≤ C + 1
1
2
for all sufficiently large |z|, whence continuity of g implies it is bounded and Liou-ville’s theorem implies it is a constant a. Thus f (z) = azn + P(z) is a polynomialof degree at most n.
4.2.3.5 Suppose f is analytic at z0. Let g(z) = f (z)− f (z0). If n ≥ 1 is an integer and r > 0is sufficiently small (so f is analytic on |z| ≤ r), then by the Cauchy formula∣∣∣ f (n)(z0)∣∣∣ = ∣∣∣g(n)(z0)
∣∣∣ = ∣∣∣∣∣∣ n!2πi
∫|z−z0 |=r
g(z)(z − z0)n+1 dz
∣∣∣∣∣∣ ≤ n!rn max|z−z0 |=r
|g(z)| =n!rn max|z−z0 |=r
| f (z)− f (z0)|.
Putting r = 1/n we see that n!nn∣∣∣ f (n)(z0)
∣∣∣ ≤ max|z−z0 |=1/n | f (z) − f (z0)| → 0 asn → ∞. This is strictly stronger than the statement that
∣∣∣ f (n)(z0)∣∣∣ > n!nn cannot
occur for an unbounded sequence of n ∈ N.4.3.2.1 The algebraic order of f at a is defined to be that h such that
limz→a|z − a|α| f (z)| =
∞ α < h0 α > h.
If f has order h and g has order k at z = a, then it is immediate that |z−a|α| f (z)g(z)| =∞ for α < h + k and = 0 for α > h + k, so the order of f (z)g(z) is h + k. It is alsoeasy to see that the order of 1/g is −k; combining these results we see that theorder of f (z)/g(z) is h − k.
To see that the order of f + g cannot exceed maxh, k it suffices to note that|z − a|α| f (z) + g(z)| ≤ |z − a|α| f (z)| + |z − a|α|g(z)| → 0 as z → a, by the triangleinequality.
4.3.2.3 The map ez takes any infinite horizontal strip of width 2π to C \ 0, so it takesany punctured neighborhood |z| > R of ∞ to C \ 0. It follows that ez does nothave a limit in C as z→ ∞, and therefore that the isolated singularity of ez at∞ isessential. Similar arguments apply to cos z and sin z. All we need verify is that theimage of |z| > R cannot converge in C as R → ∞, which is true for cos because itmaps all lines z = 2πk + iy onto [0,∞) ⊂ R and for sin because it maps the samelines onto the imaginary axis.
4.3.2.5 Suppose f has an isolated singularity at a, so f is analytic on 0 < |z − a| <δ. If either < f or = f is bounded on this punctured disc, then the image of thepunctured disc cannot be dense in C, so the Casorati-Weierstrass theorem impliesthe singularity cannot be essential. The singularity is therefore removable or apole. However, if it is a pole then f (z) = (z − a)−kg(z) for some k ∈ N and ganalytic on |z − a| < δ with g(a) , 0. Then we can easily verify that both < fand = f are unbounded on 0 < |z − a| < δ as follows. Take z = a + reiθ, sof (z) = r−ke−ikθg(z). For r > 0 so small that |g(z) − g(a)| < |g(a)|/2 and θ chosenso e−ikθg(a) is real we have < f (z) ≥ r−k |g(a)|/2. A similar argument works forthe imaginary part. It follows that the singularity cannot be a pole, so it must beremovable.
4.3.2.6 Let f have an isolated singularity at a, so f is analytic on 0 < |z− a| < δ. Considerg(z) = e f (z), which also has an isolated singularity at a. Suppose the singularity ofg(z) is a pole. Then limz→a |g(z)| = ∞, and |g(z)| = e< f (z), so limz→a< f (z) = ∞.In particular, f (z) is unbounded on 0 < |z − a| < δ, so the singularity of f is notremovable. Also, f (z) omits a left half-plane, so the Casorati-Weierstrass theoremimplies the singularity of f cannot be essential. Thus f has a pole at a. Reasoningas in the previous question we see that for any sufficiently small r > 0, the imageof any 0 < |z − a| < r under f covers a set of the form |z| > R. However the
3
exponential map takes any horizontal infinite strip of height 2π to C \ 0. Since aregion of the form |z| > R contains such a horizontal strip we conclude that g mapsany 0 < |z − a| < r to all of C \ 0, contradicting the assumption that g has a poleat a. We conclude that if f (z) has an isolated singularity at a then e f (z) cannot havea pole at a.
Math 5120: Complex analysis. Homework 7 Solutions4.3.3.1 Observe that f (0) = 0 and f ′(0) , 0, so that f is injective on some neighborhood
of 0. Also f (z) is not injective on |z| < 2, because f (0) = f (−1) = 0. It followsthat R = supr : f is injective on |z| < r is positive and finite.
One easy way to find R is by direct computation. The roots of f (z) − ζ are
z = − 12 ±
√1+4ζ2
2 , so they are at opposite points of a diameter of a circle radius
s =∣∣∣ √1+4ζ2
2
∣∣∣ around − 12 . One of these points has real part less than − 1
2 , so isdistance at least 1
2 from 0. We conclude that no open disc |z| < r with r ≤ 12
contains two roots of f (z) − ζ. Moreover if r > 12 we may choose ζ to make s so
small that |z − 12 | < s is inside |z| < r, at which point there will be two roots of
f (z) − ζ in |z| < r. This proves that R = 12 .
A more general approach is to first recognize that an analytic f cannot be in-jective on an open set containing a critical point. The reason is that if f ′(z0) = 0,then f (z) − f (z0) has a zero of some order n ≥ 2 at z0. The argument principlethen says that if w is sufficiently close to f (z0), the function f (z) − w must have nsimple roots in a neighborhood of z0, so f cannot be injective. (See book, top ofpage 132.) For our problem, the critical point is at 2z + 1 = f ′(z) = 0, so z = − 1
2and we discover R ≤ 1
2 .Now if f is a polynomial, then we know that all critical points lie in the convex
hull of the roots (see Theorem 1 on page 29 of the book). So if we have a polyno-mial f of degree n and a w so f (z)−w has n roots (counting multiplicity) in |z| < r,then f also has a critical point in |z| < r. In the situation we face, with n = 2, wediscover that if there are 2 roots of f (z) − ζ in |z| < r then r > 1
2 . Thus R ≥ 12 , and
we conclude R = 12 .
It is interesting to think how, if at all, this idea could be generalized.4.3.3.4 We are given f analytic at 0 and f ′(0) , 0. Then f (z) − f (0) = zh(z) with h
analytic at 0 and h(0) = f ′(0) , 0. We may then take an open disc U around 0which is so small that h is analytic on this disc and |h(z) − h(0)| < |h(0)| for z ∈ U.This condition on h implies log h(z) and hence k(z) = h(z)1/n are well-defined(single-valued) analytic functions on U, from which f (zn) = f (0) + znh(zn) =f (0) +
(zk(zn)
)1/n on U, so taking g(z) = zk(zn) completes the proof.
1
Math 5120: Complex analysis. Homework 8 Solutions4.3.4.1 Suppose | f (z)| ≤ 1 on |z| ≤ 1 and f analytic. Fix z0 in the open unit disc and
let w0 = f (z0). Using a composition of fractional linear transformations and theSchwarz lemma, it is proved in the book that
(1)∣∣∣∣∣ f (z) − w0
1 − w0 f (z)
∣∣∣∣∣ ≤ ∣∣∣∣∣ z − z0
1 − z0z
∣∣∣∣∣ .With a little rewriting we can extract f ′(z0):
(2)| f ′(z0)|
1 − | f (z)|2= lim
z→z0
∣∣∣∣∣ f (z) − w0
(z − z0)(1 − w0 f (z))
∣∣∣∣∣ ≤ limz→z0
∣∣∣∣∣ 11 − z0z
∣∣∣∣∣ = 11 − |z0|
2
and z0 was an arbitrary point in the disc, so the estimate is valid for all |z0| < 1.4.3.4.2 Suppose f is non-constant analytic and = f (z) ≥ 0 if =z ≥ 0. Composition with
a fractional linear map gives a map from the unit disc to itself to which we canapply the Schwarz lemma. Specifically, for any z0 with =z0 > 0 we set w0 = f (z0)and have =w0 > 0 by the maximum principle. For α in the upper half-plane letgα(z) = z−α
z−α so g−1z0
maps the unit disc to the upper half-plane with 0 7→ z0 and gw0
maps the upper half-plane to the unit disc with w0 7→ 0. Thus gw0 f g−1z0
takesthe unit disc to itself and fixes 0, whence |gw0 f g−1
z0(z)| ≤ |z| by the Schwarz
lemma, and so |gw0 f (z)| ≤ |gz0 (z)|. Substituting gives
(3)
∣∣∣∣∣∣ f (z) − f (z0)
f (z) + f (z0)
∣∣∣∣∣∣ ≤∣∣∣∣∣ z − z0
z + z0
∣∣∣∣∣ .As in the previous question, we can obtain a bound for | f ′(z0)| by dividing both
sides of the previous equation by |z − z0| and sending z → z0. Since z + z = 2=z,we obtain
(4)| f ′(z0)|= f (z0)
≤1=z0
valid at all points in the upper half-plane.4.3.4.3 Let
F(z) =( f (z) − w0)(1 − z0z)(z − z0)(1 − w0 f (z))
.
Then F is analytic on the open unit disc except for a removable singularity at z0,and (1) says |F(z)| ≤ 1. If equality holds in (2) then |F(z0)| = 1, so the maximumprinciple implies F is constant of modulus 1, which we may write as eiθ, θ ∈[0, 2π). Multiplying out gives that f (z) = h−1
w0(eiθhz0 (w)) where hα = z−α
1−αz .Similarly, if we let
G(z) =( f (z) − w0)(z + z0)(z − z0)( f (z) + w0)
then we obtain a function analytic on =z > 0 except for the removable singularityat z0, and |G(z)| ≤ 1 by (3). Equality in (4) implies G ≡ eiθ for some θ ∈ [0, 2π) bythe same maximum principle argument as before, and inverting the maps we havef (w) = g−1
w0(eiθgz0 (w)) with g as in the previous exercise.
In either of these two cases we see that f is a composition of fractional linearmaps, so is fractional linear.
4.4.7.3 Let γ be a closed curve. Its complement is open, so the connected componentsof the complement are open, and there are thus at most countably many of them.Label them U j
∞j=0, and let U0 be the unique unbounded one; it is unique because
1
2
compactness of γ implies the complement of some large disc is in C \ γ, and thisis a connected set so is contained in a single component.
For each j, let V j = C\U j and observe V j = ∪k, jU j∪γ∪∞. As these sets areconnected, each must lie in exactly one component of V j. However connectivityof Uk implies connectivity of its closure, which intersects γ, so all Uk and γ mustlie in a single component of V j. Moreover if j , 0 then U0 ⊂ V j and ∞ is itsclosure, so is in this same component of V j. We conclude that if j , 0 then V j hasonly one component, so U j is simply connected by definition. Also V0 has at mosttwo components, one being ∞ and the other ∪ j ≥ 1U j ∪ γ. Taking r so |z| = rdoes not intersect γ, we see that the disjoint open sets |z| > r and |z| < r separateV0 into these two components, so V0 is doubly connected.
4.4.7.5 Let Ω be a region not intersecting a connected set E with ±1 ∈ E. If φ(z) is aglobally analytic function such that f (z) = 1−z2
(φ(z))2 maps E to a connected regioncontaining 0, then f has a well-defined logarithm and therefore well-defined rootson on C \ f (E). Provided f (Ω) does not intersect f (E) we may define a single-valued analytic function
√1 − z2 on Ω by√1 − z2 = φ(z)
√1 − z2
(φ(z))2
and it is a legitimate branch of the square root because squaring both sides leadsto an equality.
It remains to find such a φ(z), but writing 1 − z2 = (1 − z)(1 + z) immediatelysuggests φ(z) = (1+ z), as then f (z) = 1−z
1+z , which is a fractional linear transforma-tion with 1 7→ ∞ and 1 7→ 0, from which f (E) is a connected set containing 0 and∞. Our definition is then√
1 − z2 = (1 + z)
√1 − z1 + z
.
Now consider the integral∫γ
dz√
1 − z2=
∫γ
1
(1 + z)√
1−z1+z
dz
where γ is a closed curve in a component of C \ E. The integrand has no singu-larities in this component, so if γ does not wind around any point of E then theintegral is zero. In particular if∞ ∈ E then γ cannot wind around E, so the integralis zero in this case.
We therefore suppose that E is bounded and take r > 0 so large that E ⊂ z :|z| < r. In this case the winding number n(γ, z) is a constant 2πiN on E and γ ishomologous in the unbounded component of C \ E to Nγr, where γr is the circleof radius r around 1. The Cauchy theorem then implies∫
γ
dz√
1 − z2= N∫γr
1
(1 + z)√
1−z1+z
dz. = N∫ 2π
0
ireiθ dθ
reiθ√
2reiθ − 1
= iN∫ 2π
0
dθ√2
reiθ − 1
It is tempting at this point to attempt to use the residue theorem, but that theoremis valid only for isolated singularities, and our square root has singularities alonga connected set joining ±1. Instead we observe that we may take r → ∞ withoutchanging the value of the integral. We find that the integrand converges to 1/
√−1,
which is one of ±i (we cannot determine which without knowing more about the
3
set E; draw some pictures to see why). Therefore the possible values of the integralare ±2πN, or any element of 2πZ.
4.5.2.1 Let f (z) = z7 −2z6 − z+1. Then | f (z)| ≤ 5 on |z| = 1, so Rouche’s theorem implies6z3 + f (z) has the same number of roots as 6z3 in the unit disc, namely three.
4.5.2.2 We use Rouche’s theorem twice. For |z| = 2, |z4| = 16 > | − 6z + 3|, so z4 − 6z + 3has 4 roots in |z| ≤ 2. For |z| = 1, |6z| = 6 > |z4 + 3|, so there is one root in |z| ≤ 1.We conclude that there are 3 roots in 1 < |z| < 2.
Math 5120: Complex analysis. Homework 9 Solutions4.5.3.1.a
f (z) =1
z2 + 5z + 6=
1(z + 2)(z + 3)
which has• a pole of order 1 at z = −2 with residue limz→−2(z + 2) f (z) = 1• a pole of order 1 at z = −3 with residue limz→−3(z + 3) f (z) = −1
4.5.3.1.b
f (z) =1
(z2 − 1)2 =1
(z − 1)2(z + 1)2
which has• a pole of order 2 at z = 1 with residue limz→1
ddz (z − 1)2 f (z) = − 1
4• a pole of order 2 at z = −1 with residue limz→−1
ddz (z + 1)2 f (z) = 1
44.5.3.1.c f (z) = 1
sin z has poles at the zeros of sin z, so at the points πk, k ∈ Z. These zerosare simple, because f ′(z) = cos z = ±1 at these points, and consequently the polesare simple. The residue at πk may be computed by L’Hopital’s rule
limz→πk
z − πksin z
= limz→πk
1cos z
= (−1)k
so that f (z) has simple poles with residue (−1)k at each πk, k ∈ Z.4.5.3.1.d As cos z is entire, f (z) = cot z = cos z
sin z can only have poles at the zeros of sin z,meaning the points z = πk, k ∈ Z. Since cos z , 0 at these points, there is a pole ateach such point, and since the zeros of sin z are simple the poles are also simple.The residue at πk is
limz→πk
(z − πk) cos zsin z
= cos(πk) limz→πk
1cos z
= 1
so cot z has simple poles with residue 1 at each πk, k ∈ Z.4.5.3.1.e f (z) = 1
sin2 zhas poles at each of the zeros z = πk, k ∈ Z of sin z. These zeros are
order 1, so the zeros of sin2 z are order 2. The residues may be computed usingL’Hopital
limz→πk
ddz
( (z − πk)2
sin2 z
)= lim
z→πk
2(z − πk) sin z − 2(z − πk)2 cos z
sin3 z
= 2 limz→πk
( z − πksin z
)limz→πk
( sin z − (z − πk) cos z
sin2 z
)= 2 lim
z→πk
(cos z − cos z + (z − πk) sin z2 sin z
)= lim
z→πk(z − πk) = 0.
We determine that f (z) has poles of order 2 at each of the points πk, k ∈ Z and hasresidue zero at each pole.
4.5.3.1.f f (z) = z−m(1 − z)−n, m, n ∈ N has a pole of order m at 0 and a pole of order n at 1.We may compute the residue at 0 using
limz→0
1(m − 1)!
dm−1
dzm−1 (1 − z)−n =1
(m − 1)!(n + m − 2)!
(n − 1)!(1 − 0)−n−(m−1)
=
((n − 1) + (m − 1)
m − 1
)=
((n − 1) + (m − 1)
n − 1
)1
2
The residue at 1 may be computed the same way
limz→1
1(n − 1)!
dn−1
dzn−1 (z − 1)n f (z) =1
(n − 1)!limz→1
dn−1
dzn−1 (−1)nz−m
= (−1)n(−1)n−1 (m + n − 2)!(n − 1)!(m − 1)!
(1)−m−(n−1)
= −
((n − 1) + (m − 1)
m − 1
)= −
((n − 1) + (m − 1)
n − 1
)so we conclude that f has a pole of order m at zero with residue
((n−1)+(m−1)
n−1
)and
a pole of order n at 1 with residue −(
(n−1)+(m−1)n−1
). It is worth noting that this is
consistent with exercise 4.2.3.1(b).4.5.3.3 There are a few things we will use repeatedly in computing these integrals. The
curve γR will be the semicircle |z| = R in the upper half-plane, with the usual(increasing angle) orientation. For R, S ,T ∈ (0,∞) we also let Γ1 = S + iy : 0 ≤y ≤ T , Γ2 = x + iT : −R ≤ x ≤ S , Γ3 = −R + iy : 0 ≤ y ≤ T , oriented suchthat Γ1 + Γ2 + Γ3 and the interval [−R, S ] ⊂ R form a positively oriented closedcurve. We will frequently use that (A) if the integrand f (x) is bounded by |x|−2 as|x| → ∞ then
∫ ∞−∞= limR→∞
∫ R−R f (x), and (B) if the integrand f (z) is bounded by
|z|−2 as |z| → ∞ then limR→∞∫γR
f (z)dz = 0
4.5.3.3.a Note that sin2 x = sin2(−x) = sin2(π − x) = sin2(π + x) implies∫ π/2
0
dx
a + sin2 x=
14
∫ 2π
0
dx
a + sin2 x
and this may be seen as an integral on the unit circle with respect to the angledx = dz/iz where sin z = (z − z−1)/2i. Thus∫ π/2
0
dx
a + sin2 x=
14
∫|z|=1
1
a +((z − z−1)/2i
)2
dziz.
We simplify((z− z−1)/2i
)2= −(2z)−2(z4 − 2z2 + 1) and find the integrand becomes
4izz4−(2+4a)z2+1 . At this point we can make our lives a little easier by making thesubstitution w = z2. Notice that when z winds once around the unit circle, w windsaround twice. We therefore find∫ π/2
0
dx
a + sin2 x=
12
∫|w|=1
2iw2 − (2 + 4a)w + 1
dw = i∫|w|=1
dww2 − (2 + 4a)w + 1
.
Now we would like to say that the poles of w2 − (2 + 4a)w + 1 are at w± =1+2a±2
√a2 + a by the quadratic formula, but this requires that we make sense of
the square root. Fortunately, |a| > 1 by hypothesis, so |1/a| < 1 and√
1 + (1/a) iswell-defined. We may therefore define an analytic branch of
√a2 + a by
√a2 + a =
a√
1 + (1/a), obtaining w± = 1 + 2a ± 2a√
1 + (1/a). By construction, each ofw± is a branch of the inverse of the map w 7→ (w + w−1)2/4, evaluated at a. Sincew 7→ (w + w−1)2/4 takes the unit circle to the interval [−1, 1], and |a| > 1, we seethat |w±| , 1 (this is important for applying the residue theorem). It also followsthat only one of w± can lie inside the unit disc, and that which one does so is in-dependent of a. Taking a ∈ (1,∞) we readily see |w−| < 1 and |w+| > 1, so thismust be true for all a. Thus the result of the integration can be computed from the
3
residue at the simple pole w−, which has value 1/(w−−w+) = −1/(4a√
1 + (1/a)).Finally
∫ π/2
0
dx
a + sin2 x=
2πi2
−(4a√
1 + (1/a))=
π
2a√
1 + (1/a).
4.5.3.3.b Using that the integrand is even and (A), then the residue theorem, and then (B)
∫ ∞
0
x2 dxx4 + 5x2 + 6
=12
∫ ∞
−∞
z2 dzz4 + 5z2 + 6
=12
limR→∞
∫ R
−R
z2 dzz4 + 5z2 + 6
= πi∑
j
Resz j −12
limR→∞
∫γR
z2 dzz4 + 5z2 + 6
= πi∑
j
Resz j
z2
z4 + 5z2 + 6
where the sum is over residues in the upper half-plane. Now z4 + 5z2 + 6 =(z2 + 2)(z2 + 3), so the integrand has simple poles at z = ±i
√2 and z = ±i
√3. The
residue at i√
2 is −2/(2√
2i)(1), and at i√
3 is −3/(−1)(2√
3i), so the result is
∫ ∞
0
x2 dxx4 + 5x2 + 6
= πi(−√2
2i+
√3
2i
)= (√
3 −√
2)π
2.
4.5.3.3.c By (A), the residue theorem, and (B)
∫ ∞
−∞
x2 − x + 2x4 + 10x2 + 9
dx
= limR→∞
∫ R
−R
z2 − z + 2z4 + 10z2 + 9
dz
= 2πi∑
j
Resz j − limR→∞
∫γR
z2 − z + 2z4 + 10z2 + 9
dz
= 2πi∑
j
Resz j
z2 − z + 2z4 + 10z2 + 9
where the sum is over residues in the upper half-plane. The zeros of z4+10z2+9 =(z2 + 1)(z2 + 9) are at ±i and ±3i; each produces a simple pole in the integrand,and there are no others. The residue at i is (−1 − i + 2)/(2i)(−1 + 9) and at 3i is(−9 − 3i + 2)/(−9 + 1)(6i), so the result is
∫ ∞
−∞
x2 − x + 2x4 + 10x2 + 9
dx = 2πi(1 − i
16i+
7 + 3i48i
)= (3 − 3i + 7 + 3i)
π
24=
5π12.
4
4.5.3.3.d Using that the integrand is even and (A), then the residue theorem, and then (B)∫ ∞
0
x2 dx(x2 + a2)3 =
12
∫ ∞
−∞
z2 dz(z2 + a2)3
=12
limR→∞
∫ R
−R
z2 dz(z2 + a2)3
= πi∑
j
Resz j −12
limR→∞
∫γR
z2 dz(z2 + a2)3
= πi∑
j
Resz j
z2
(z2 + a2)3
where the sum is over residues in the upper half-plane. Factoring (z2 + a2)3 =
(z + ai)3(z − ai)3, a ∈ R, we see that there is a single pole of order 3 in the upperhalf-plane, at i|a|. The residue there is
limz→|a|i
d2
dz2
z2
(z + |a|i)3 =1
8|a|3i
so that the result is ∫ ∞
0
x2 dx(x2 + a2)3 =
π
8|a|3
4.5.3.3.e Using that the integrand is even and (A), that cos z = <eiz, then the residue theo-rem for R = S and T sufficiently large,∫ ∞
0
cos x dxx2 + a2 =
12
∫ ∞
−∞
cos z dzz2 + a2
=12
limR→∞<
∫ R
−R
eiz dzz2 + a2
= πi∑
j
Resz j
eiz
z2 + a2 −12
limR,T→∞
<
(∫Γ1
eiz dzz2 + a2
∫Γ2
eiz dzz2 + a2 +
∫Γ3
eiz dzz2 + a2
)
where the sum is over residues in the upper half-plane. However the integrand isbounded by a constant multiple of e−y/|z|2 for z = x + iy and |z| sufficiently large.Writing f (z) for the integrand, and taking R = S and T large enough we find that∣∣∣∫Γ1
f (z) dz∣∣∣ ≤ R−2
∫ T0 e−y dy ≤ R−2, and similarly for Γ3. Now on Γ2 we have that
(z2+a2)−1 is integrable (with integral bounded by constant C) if T is large enough,and therefore
∣∣∣∫Γ2
f (z) dz∣∣∣ ≤ Ce−T . Sending R and T to∞ we find∫ ∞
0
cos x dxx2 + a2 = <πi Res|a|i f (z) =
π
2|a|
where at the last step we computed that z2 + a2 = (z + ai)(z − ai), has one simplepole in the upper half-plane, at i|a|, with residue limz→|a|i
cos zz+|a|i =
cos a2|a|i .
4.5.3.3.f We use that the integrand is even and x sin x/(x2 + a2) = =zeiz/(z2 + a2). TakingR, S ,T large enough that the curve (−R, S )∪Γ1 ∪Γ2 ∪Γ3 encloses the simple pole
5
at |a|i, where the residue is limz→|a|izeiz
z+|a|i =|a|ie−|a|
2|a|i =e−|a|
2 we obtain∫ ∞
0
x sin xx2 + a2 dx =
12
limR,S→∞
=
∫ S
−R
zeiz
z2 + a2 dz
= =πie−|a|
2− lim
R,S→∞=
(∫Γ1
zeiz
z2 + a2 dz +∫Γ2
zeiz
z2 + a2 dz +∫Γ3
zeiz
z2 + a2 dz)
valid for all sufficiently large T . However, the integrand f (z) satisfies | f (z)| ≤|z|e−y
|z|2−|a|2 for z = x+iy. The integral for Γ1 can be bounded by RR2−|a|2
∫ T0 e−ydy = R
R2−|a|2
and similarly that for Γ3 can be bounded by SS 2−|a|2 . The integral for Γ2 can be
bounded by S e−T
S 2−|a|2 (R + S ). If we first send T → ∞ so the Γ2 integral goes to 0,and then send R, S → ∞ we find that they make no contribution to the result, andtherefore ∫ ∞
0
x sin xx2 + a2 dx = =πi
e−|a|
2=π
2e|a|
4.5.3.3.g We will do this for general β ∈ (−1, 1), as it will be useful later. Take δ > 0, ε > 0,and R > 2. Let Γ± = rei±δ, r ∈ (ε,R) be rays at angle ±δ, and also take arcsΓR = Reiθ : θ ∈ (δ, 2π − δ) and Γε = εeiθ : θ ∈ (δ, 2π − δ). Let zβ be a branch onC \ [0,∞), so it is well-defined and analytic in a simply connected neighborhoodof the closed curve Γ+ + ΓR − Γ− − Γε . Provided δ and ε are sufficiently small, thiscurve winds once around the simple poles of f (z) = zβ(1 + z2)−1, which are at ±i,and where there are residues iβ/2i and (−i)β/ − 2i respectively.∫
Γ++ΓR−Γ−−Γε
zβ
z2 + 1dz = 2πi(iβ − (−i)β)/2i = π(eiπβ/2 − ei3πβ/2).
Now on Γ+ we have zβ = rβeiδβ, while on Γ−, zβ = rβei(2π−δ)β. It follows that
lim δ→ 0∫Γ+−Γ−
zβ
z2 + 1dz = (1 − ei2πβ)
∫ R
ε
xβ
x2 + 1dx.
At the same time, we see that on Γε the integrand has the bound | f (z)| ≤ 2εβ, andthe length of the curve is less than 2πε, so the integral is bounded by 4πε1+β → 0as ε → 0, provided β > −1. On ΓR we have | f (z)| ≤ Rβ/(R2 − 1), and the curvehas length less than 2πR, so the integral is bounded by 2πR1+β/(R2 − 1) → 0 asR→ ∞ provided β < 1. We conclude that if β ∈ (−1, 1) then∫ ∞
0
xβ
x2 + 1dx = lim
ε→0,R→∞,δ→0
11 − ei2πβ
∫Γ++ΓR−Γ−−Γε
zβ
z2 + 1dz
= πeiπβ/2 − ei3πβ/2
1 − ei2πβ
= πeiπβ(e−iπβ/2 − eiπβ/2)
eiπβ(e−iπβ − eiπβ)
= πsin(πβ/2)
sin πβ= π
sin(πβ/2)2 sin(πβ/2) cos(πβ/2)
=π
2sec
(πβ2
)In the special case β = 1
3 , we have sin(π/6)/ sin(π/3) = 1/√
3, so that∫ ∞
0
x1/3
x2 + 1dx =
π√
3.
6
4.5.3.3.h For this problem, let Γ+ = (ε,R) ⊂ R and Γ− = (−R,−ε) ⊂ R, Γε and ΓR be thesemicircles of radius ε and R (respectively) in the upper half-plane. Define log z tobe the branch of the logarithm on the complement of the negative imaginary axis.Taking Γ++ΓR−Γ−−Γε arranged to form a curve winding once around the simplepole of log z
(z2+1) at z = i, we find from the residue theorem that∫Γ++ΓR+Γ−−Γε
log zz2 + 1
dz = 2πi(log i)/2i =iπ2
2.
The computations showing that the contributions from Γε and ΓR vanish in thelimit are essentially the same as in exercise 4.5.3.3.g. All that is different is we usethe bound | log z| ≤ (log |z| + 2π). Since log z is log |z| on Γ+ and log |z| + πi on Γ−we find∫
Γ++Γ−
log zz2 + 1
dz =∫ R
ε
log xx2 + 1
dx+∫ −ε
−R
log x + πix2 + 1
dx = 2∫ R
ε
log xx2 + 1
dx+∫ R
ε
1x2 + 1
dx.
Combining these facts we see
iπ2
2= limε→0,R→∞
∫Γ++ΓR+Γ−−Γε
log zz2 + 1
dz
= limε→0,R→∞
∫Γ++Γ−
log zz2 + 1
dz
= 2∫ ∞
0
log xx2 + 1
dx +∫ ∞
0
iπx2 + 1
dx
= 2∫ ∞
0
log xx2 + 1
dx +iπ2
2
so that ∫ ∞
0
log xx2 + 1
dx = 0.
4.5.3.3.i Let us first observe that f (x) = x(−1−α) log(1 + x2) is integrable on [0,∞) becauseit is bounded by Cαx−1−(α/2) as x→ ∞ and α > 0, while as x ↓ 0 one has | log(1 +x2)| ≤ 2x2 so | f (x)| ≤ x1−α and α < 2. It follows that we can write the integral as alimit and can integrate by parts∫ ∞
0
log(1 + x2)x1+α dx = lim
R→∞
∫ R
1R
log(1 + x2)x1+α dx = lim
R→∞
[x−α
−αlog(1 + x2)
]R
1R
+1α
limR→∞
∫ R
1R
2x1−α
1 + x2 dx.
We observe that as R → ∞, R−α log(1 + R2) → 0, and also |Rα log(1 + R−2)| ≤2R2−α → 0, so the boundary term from the integration makes no contribution inthe limit. The remaining term may be dealt with by the computation in 4.5.3.3.g.Indeed, from that problem with β = (1 − α) ∈ (−1, 1), we have∫ ∞
0
log(1 + x2)x1+α dx =
1α
∫ ∞
0
2x1−α
1 + x2 dx =π
αsec
( (1 − α)π2
)=π
αcsc
(απ2
)4.5.3.4 Parameterizing |z| = ρ by z = ρeiθ we have dz = izdθ and |dz| = ρdθ, so |dz| =
ρdz/iz. Also |z − a|2 = (z − a)(z − a) = (z − a)( ρ2
z − a). Hence we find∫|z|=ρ
|dz||z − a|2
=
∫|z|=ρ
ρ
iz(z − a)( ρ2
z − a)dz =
∫|z|=ρ
ρ
i(z − a)(ρ2 − az)dz
7
which can be computed by the residue theorem. There are simple poles at a andρ2/a. By hypothesis, |a| , ρ; if |a| < ρ then a is inside |z| = ρ and ρ2/a is not, andthe reverse is true if |a| > ρ.
The residue at z = a is ρi(ρ2−|a|2) and that at ρ2/a is −ρ
i(ρ2−|a|2) . We conclude fromthe residue theorem that∫
|z|=ρ
|dz||z − a|2
=
2πρρ2−|a|2 if |a| < ρ−2πρρ2−|a|2 if |a| > ρ
=2πρ∣∣∣ρ2 − |a|2
∣∣∣ .
Math 5120: Complex analysis. Homework 10 Solutions4.6.2.2 If M(r) = 0 for some r > 0 then f vanishes on a set containing a limit point, so
f ≡ 0 and the result is trivial. Hence there is no loss of generality in assumingM(r) > 0 for r > 0, in which case the statement
M(r) ≤ M(r1)αM(r2)(1−α)
for α = log(r2/r)log(r2/r1) is equivalent to
log M(r) ≤ α log M(r1)+(1−α) log M(r2) =log r2 − log rlog r2 − log r1
log M(r1)+log r − log r1
log r2 − log r1log M(r2)
which is the same as(log r2− log r1
)log M(r) ≤ log r2 log M(r1)− log r1 log M(r2)+
(log M(r2)− log M(r1)
)log r
or(log M(r1)− log M(r2)
)log r+
(log r2− log r1
)log M(r) ≤ log r2 log M(r1)− log r1 log M(r2)
and it is this that we will prove.It is suggested in the book that we apply the maximum principle (for harmonic
functions) to a linear combination of log |z| + log | f (z)|. Of course we cannot dothis directly if f has zeros, because log | f (z)| is not harmonic in any neighborhoodof a zero of f (in fact it is subharmonic, and there is still a maximum principle forsubharmonic functions, but we have not proved that). We will therefore need to dosomething about points where f is zero, but let us begin by assuming that no suchpoints exist.
If f is analytic on the annulus 0 < r1 < |z| < r2 then A log |z| + B log | f (z)| isharmonic there, and the maximum principle for harmonic functions implies thatthe maximum occurs on the boundary. We obtain
(1)A log r+B log M(r) = max
|z|=r
(A log |z|+B log | f (z)|
)≤ maxA log r1+B log M(r1), A log r2+B log M(r2)
Taking A = log M(r1)− log M(r2) and B = log r2 − log r1 we find that the terms onthe right are both equal to log r2 log M(r1) − log r1 log M(r2). Thus
log r2 log M(r1)− log r1 log M(r2) ≥(log M(r1)− log M(r2)
)log r+
(log r2− log r1
)log M(r)
which is what we needed to prove.Now we deal with the points z j where f (z) = 0. Such points are can accumu-
late only at the boundary. Suppose that around each we place a small disc of radiusδ j (small enough that it is inside the annulus), and delete these discs from our do-main. Then (1) must be modified so that for each j there is a term on the rightside corresponding to the maximum of A log |z| + B log | f (z)| on the new boundarycircle |z − z j| = δ j. However B log | f (z)| → −∞ as z→ z j, so we may choose δ j sosmall that this new term is less than the right side of (1), and therefore need not beincluded. It follows that (1) is still valid when f has zeros, and therefore the resultholds for general f .
Note that there is a degenerate case we did not consider, namely r1 = 0. Inthis situation one should interpret the formula for α as corresponding to α = 0,whereupon the result follows directly from the usual maximum principle for theharmonic function | f (z)|.
1
2
5.1.1.1 Let K ⊂ C be compact and M = maxz∈K |z|. Observe that for n > M we have∣∣∣ zn
∣∣∣ < 1 when z ∈ K. The principal branch of the logarithm is well-defined onw : |1 + w| < 1, so we conclude log
(1 + z
n)
is well-defined on K for all n > Mand has Taylor expansion
log(1 +
zn
)=
∞∑j=1
(−1) j+1
j
( zn
) j.
The series is readily seen to be convergent on∣∣∣1+ z
n
∣∣∣ < 1, thus uniformly convergenton compact subsets of this region, and in particular on K for n > M. Uniformityof the convergence implies we can exchange the limits in
limn→∞
n log(1 +
zn
)=
∞∑j=1
limn→∞
(−1) j+1
jz j
n j−1 = z
Exponentiating both sides and using continuity of the exponential we get that
ez = limn→∞
exp n log(1 +
zn
)= lim
n→∞
(1 +
zn
)n
uniformly on K, and since K was arbitrary the convergence is uniform on all com-pact sets in C.
5.1.2.3 We wish to develop log( sin z
z)
around 0 up to terms of order z6. Since sin z has asimple zero at 0 the function sin z
z has a removable singularity at 0 and its extension(which is equal to 1 at 0) is entire. It is helpful to recall the series for sin z anddivide by z to obtain a series convergent uniformly on all compact sets to sin z
z
sin zz=
1z
∞∑j=0
(−1) jz2 j+1
(2 j + 1)!=
∞∑j=0
(−1) jz2 j
(2 j + 1)!= 1 +
∞∑j=1
(−1) jz2 j
(2 j + 1)!.
Next we may compose with the series∑∞
k=1(−1)k+1
k wk for log(1 + w), which is con-vergent for |w| < 1. This amounts to setting w =
∑∞j=1
(−1) jz2 j
(2 j+1)! , which we notesatisfies |w| < 1 on a neighborhood of 0. Observe that since the lead z-term is z2 itsuffices to consider the 3rd-order polynomial in w. We have
log( sin z
z
)=
( sin zz− 1
)−
12
( sin zz− 1
)2+
13
( sin zz− 1
)3+ [z8]
=
3∑j=1
(−1) jz2 j
(2 j + 1)!−
12
2∑j=1
(−1) jz2 j
(2 j + 1)!
2
+13
1∑j=1
(−1) jz2 j
(2 j + 1)!
3
+ [z8]
= −z2
3!+
z4
5!−
z6
7!−
12
(−
z2
3!+
z4
5!
)2
+13
(−
z2
3!
)3
+ [z8]
=(−
13!
)z2 +
( 15!−
12(3!)2
)z4 +
(−
17!+
1(3!)(5!)
−1
3(3!)3
)z6 + [z8]
= −z3
3!+
(3 − 5)z4
23325+
(−9 + 63 − 70)z6
24345 · 7+ [z8]
= −z3
2 · 3−
z4
22325−
z6
345 · 7+ [z8]