mat 205e – theory of complex functionsmat 205e – homework 2 due 01.10.2014 1. consider a...

MAT 205E – Theory of Complex Functions

Fall 2014

Instructor : İlker BayramEEB [email protected]

Class Meets : 13.30 – 16.30, WednesdayEEB 5204

O�ce Hours : 10.00 – 12.00, Monday

Textbook : E. B. Sa↵, A. D. Snider, ‘Fundamentals of Complex Analysis’, 3rd Edition, Pearson.Supp. Text : J. W. Brown, R. V. Churchhill, ‘Complex Variables and Applications’.

Grading : 2 Midterms (30% each), Final (40%).

Webpage : There’s a ’Ninova’ page. Please log in and check.

Tentative Course Outline

• Complex NumbersBasic properties, polar representation, the notion of a domain.

• Analytic FunctionsLimit in the complex plane, continuity of a complex valued function, analyticity, Cauchy-Riemannconditions, harmonic functions.

• Basic FunctionsRational functions, exponential and trigonometric functions, the logarithm, the notion of branch, thepower function.

• Complex IntegrationContours and integrals on contours, Cauchy-Goursat theorem, Cauchy’s Integral Formula, the funda-mental theorem of algebra.

• Series RepresentationsTaylor series, Laurent series, zeros and singularities.

• Residue TheoryThe residue theorem, integrals involving trigonometric functions, indented paths, integration along abranch cut.

• MiscellaneousThe complex gradient, functions of a matrix.

MAT 205E – Homework 1

Due 24.09.2014

1. For an integer n, suppose zn = 1 but z 6= 1. Show that zn�1 + zn�2 + . . .+ z + 1 = 0.

Solution. Observe that

(z � 1)| {z }P (z)

(zn�1 + zn�2 + . . .+ z + 1)| {z }Q(z)

= zn � 1.

Now if zn = 1 then P (z)Q(z) = 0, but since z 6= 1, we have P (z) 6= 0, therefore Q(z) = 0.

2. Suppose zn and un are two convergent complex-valued sequences and their limits are zand u respectively. Using the definition of limit given in class, show that tn = zn un isalso convergent and the limit is t = z u.

Solution. Suppose ✏ > 0 is given. We need to find N such that if n � N , then |tn�t| ✏.Now since zn is convergent, for any given ✏1 > 0, we can find N1 (possibly depending on✏1) such that if n � N1, then |zn � z| < ✏1.Also, since un is convergent, for any given ✏2 > 0, we can find N2 (possibly depending on✏2) such that if n � N2, then |un � u| < ✏2 (we will specify ✏1 and ✏2 below).Therefore, if n � max(N1, N2), we have

|z u� zn un| = |z u� zn u+ zn u� zn un| |z u� zn u|+ |zn u� zn un| |u|✏1 + |zn| ✏2 |u|✏1 + |zn � z + z| ✏2 |u|✏1 + |zn � z|✏2 + |z| ✏2 |u|✏1 + ✏1 ✏2 + |z| ✏2 (1)

We want to make the sum of the three terms less than ✏ by a careful choice of ✏1 and ✏2, in away that is independent of the sequences zn and un. For that, if we let M = max{|u|, |z|},and set

✏1 = ✏2 =p✏+M2 �M, (2)

then the sum of the terms on the right hand side of (1) is less than ✏. Now we go backand choose N1, N2 accordingly. Then for N = max(N1, N2), if n � N , then |tn � t| < ✏.

3. Using the definition of continuity, show that the following functions are continuous.

(a) f(z) = z2.

(b) g(z) = z̄.

(Note that g(z) is continuous but not analytic.)

Solution. (a) Let ✏ > 0 be given. Let us try to find � > 0 such that if |z � z0| �, then|f(z)� f(z0)| ✏. This will imply that f is continuous at z0.Now if |z � z0| < �, then z = z0 + u with |u| < � (we will specify � below). Then,

|f(z)� f(z0)| = |(z0 + u)2 � z20 | = |u2 + 2 z0 u| < |u|2 + 2 |z0| |u| < �2 + 2 |z0| �

Now if

0 < � < �|z0|+p

|z0|2 + ✏, (3)

then �2 + 2 |z0| � < ✏. Thus, for instance if � =�p

|z0|2 + ✏ � |z0|�/2, we have

|f(z)� f(z0)| < ✏.Since z0 was arbitrary, it follows that f(z) is continuous everywhere.

(b) Let ✏ > 0 be given. Suppose also that |z � z0| < �, with � = ✏.

|g(z)� g(z0)| = |z̄ � z̄0| = |z � z0| < ✏.

Thus g is continuous at z0. Since z0 is arbitrary, it follows that g(z) is continuouseverywhere.

4. We noted in class that if f and g are continuous at z, then h = f g is continuous at z.Prove this, using the definition of continuity.

Solution. This is similar to Q2.

Since f is continuous, for a given ✏1, we can find �1 such that if |z � z0| < �1, then|f(z)� f(z0)| < ✏1.Similarly, since g is continuous, for a given ✏2, we can find �2 such that if |z � z0| < �2,then |g(z)� g(z0)| < ✏2.Now let � = min(�1, �2). If |z � z0| < �, we have,

|f(z0)g(z0)� f(z) g(z)| |f(z0)g(z0)� f(z0) g(z)|+ |f(z0)g(z)� f(z) g(z)| |f(z0)| ✏2 + |g(z)| ✏1 |f(z0)| ✏2 + |g(z)� g(z0) + g(z0)| ✏1 |f(z0)| ✏2 + ✏2 ✏1 + |g(z0)| ✏1

Now for M = max{|f(z0), |g(z0)||}, set

✏1 = ✏2 =p✏+M2 �M.

Go back and choose �1, �2 accordingly and set � = min(�1, �2). Then if |z � z0| < �, wehave,

|f(z0)g(z0)� f(z) g(z)| < ✏.

Thus h is continuous at z0. Since z0 is an arbitrary point, it follows that h is continuouseverywhere.

5. Suppose f(z) = zn, where n is an integer. Show that f 0(z) = n zn�1.

Solution. We will show the claim above by induction. Let Pn(z) = zn. Observe thatP 01(z) = 1 so the claim holds for n = 1. Suppose that P

0n(z) = n z

n�1. Noting that

Pn+1(z) = Pn(z)P1(z), we have,

Pn+1(z) = P0n(z)P1(z) + Pn(z)P

01(z)

= n zn�1 z + zn

= (n+ 1) zn.

Thus the claim is valid for n + 1 too. Since it is true for n = 1, by induction, it musttherefore be valid for all integers n.

2

6. Suppose f , g, h are functions which are all di↵erentiable at z0. Let d(z) = f(z) g(z)h(z).Show that d0(z0) = f 0(z0) g(z0)h(z0) + f(z0) g0(z0)h(z0) + f(z0) g(z0)h0(z0).

Solution. Let u(z) = g(z)h(z). Note that u0(z0) = g0(z0)h(z0) + g(z0)h0(z0). Thus,

d0(z0) = f0(z0) u(z0) + f(z0) u

0(z0)

= f 0(z0) g(z)h(z) + f(z0) g0(z0)h(z0) + f(z0) g(z0)h

0(z0).

3


Due 01.10.2014

1. Consider a second-order polynomial of the form

P (x, y) = a0x2 + a1x y + a2y

2,

where ai’s are possibly complex valued. Suppose P (x, y) satisfies a Cauchy-Riemann conditionof the form Py = iPx. Show that actually P (x, y) = a0(x+ iy)2.

Solution. Using Py = iPx, we find

Py = a1 x+ 2a2 y = i Px = i2 a0 x+ ia1 y.

Since this is valid for all values of x and y, we must have,

a1 = 2i a0, 2a2 = i a1.

Combining we have,

P (x, y) = a0x2 + a02ixy � a0y2 = a0 (x+ iy)2.

2. Let

u(x, y) = x2 � x� y2.

Find a function v(x, y) such that f(x, y) = u(x, y) + iv(x, y) is an entire function (where z =x+ iy).

Solution. Note that u is a polynomial so it and its partial derivatives are continous. Thefunction v together with u should satisfy the Cauchy-Riemann conditions :

ux = 2x� 1 = vyuy = �2y = �vx.

Integratiing these two conditions, we find that

v = 2xy � y + h(x) = 2xy + g(y),

so that h(x) � g(y) = y. This is only possible if g(y) = �y + c and h(x) = c for a constant c.Thus v(x, y) is of the form v(x, y) = 2xy � y + c. So we find,

f(x, y) = x2 � x� y2 + i(2xy � y + c)= (x+ iy)2 � (x+ iy) + ic= z2 � z + ic.

3. (From our textbook)

(a) Show that if f(z) is analytic and real-valued in a domain D (recall that domain is an openand connected set), then f(z) is constant throughout D.

(b) Show that if both f(z) and f(z) is analytic in a domain D, then f(z) is constant throughoutD.

(c) Show that if both f(z) and |f(z)| is analytic in a domainD, then f(z) is constant throughoutD.

Solution. (a) Let f(x, y) = u(x, y)+ iv(x, y), where u and v are real valued. If f is real valued,v = 0. But then by the Cauchy-Riemann conditions, we have, ux = vy = 0, uy = �vx = 0.Therefore, u must be a constant. Thus f(z) is a constant.

(b) Let g(z) = (f(z)� f(z))/(2i). Then, g(z) is real-valued and since it is a linear combinationof analytic functions, it is also analytic. But then by part (a), it follows that g is a constant.But g(z) is equal to the imaginary part of f(z). Let h(z) = f(z) � i g(z). Then, h isreal-valued and analytic since both f and g are analytic. But again by part (a), it must beconstant. Finally we have f(z) = h(z) + ig(z) is a constant function because both h and gare constant functions.

(c) Note that |f(z)| is real valued, so by part (a), it must be a constant.First note that if |f(z)| = 0, we must have f(z) = 0 and the claim follows trivially in thiscase.

Suppose now that |f(z)| = c 6= 0, so that f(z) 6= 0 on D. Then |f(z)|2 = c2. But|f(z)|2 = f(z) f(z) and since f(z) is analytic and non-zero on D, f(z) = c2/f(z) is alsoanalytic. The claim now follows by part (b).

4. Let

P1(z) = (z � 2)2,P2(z) = (z � 3) z.

Find two polynomials Q1(z), Q2(z) such that

P1(z)Q1(z) + P2(z)Q2(z) = 1.

Solution. Expanding the polynomials, we have,

P1(z) = z2 � 4z + 4

P2(z) = z2 � 3z.

By polynomial division, we find

P1(z) = P2(z) + (�z + 4)| {z }A1(z)

equivalently A1 = P1 � P2.

Then,

P2(z) = A1(z) (�z � 1)| {z }A2(z)

+4.

Thus,

1 =1

4

⇣P2 �A1A2

⌘

=1

4

⇣P2 � (P1 � P2)A2

⌘

= P11

4(�A2)

| {z }Q1

+P21

4(1 +A2)

| {z }Q2

5. Suppose a degree-k polynomial P (z) satisfies P (z) 6= 0 if |z| r for some r. Then it can beshown that we can find positive numbers 0 < c1 < c2 such that c1 < |P (z)| < c2 if |z| r.Use the fact above to prove the following. Suppose Q(z) is a degree n polynomial and

Q(z0) = 0,

Q0(z0) = 0,

Q00(z0) 6= 0.

2

Show that we can find ✏ > 0 and two constants 0 < d1 < d2 such that

d1|z � z0|2 < |Q(z)| < d2|z � z0|2, if 0 < |z � z0| < ✏.

Solution. Since Q(z0) =, we must have that Q(z) = (z � z0)Q1(z) where Q1 is a polynomialof degree n� 1. But then,

Q0(z) = Q1(z) + (z � z0)Q01(z),

so that

Q0(z0) = Q1(z0) = 0.

Therefore, Q1(z) = (z�z0)Q2(z), so that Q(z) = (z�z0)2Q2(z). Di↵erentiating twice, we find,

Q00(z) = 2Q2(z) + 2(z � z0)Q02(z) + (z � z0)2Q2(z),

so that

Q00(z0) = 2Q2(z0) 6= 0.

Thus z0 is not a zero of Q2(z). Now let P (z) = Q(z + z0). P is also a polynomial of degreen and P (z) = z2 P1(z), where P1(z) = Q2(z + z0). Note that P1(0) = Q2(z0) 6= 0. Therefore,by the fact stated in the question statement above, we can find r such that if 0 < |z| < r, thenthere are constants c1, c2 such that c1 < |P1(z)| < c2. But this implies that

|P (z)| = |z2| |P1(z)| < c2|z2|,|P (z)| = |z2| |P1(z)| > c1|z2|,

if 0 < |z| < r. Combining,

c1|z|2 < |P (z)| < c2|z|2 if 0 < |z| < r.

Let t = z + z0. Then we can write the inequalities above as,

c1|t� z0|2 < |P (t� z0)| {z }Q(t)

| < c2|t� z0|2 if 0 < |t� z0| < r.

3


Due 08.10.2014

1. Suppose f(z) is entire. Show that g(z) = f(z̄) is also entire. (Notice that f(z) = g(z̄).)

Solution. Let f(x, y) = u(x, y)+ iv(x, y) where z = x+ iy and such that x, y, u, v are allreal valued. Then, if h(z) = f(z̄), we can write h(x, y) = f(x,�y) = u(x,�y)+ iv(x,�y).Finally, if g(z) = g(z) = f(z̄) = h(z), then

g(x, y) = u(x,�y)| {z }q(x,y)

+i��v(x,�y)

�| {z }

w(x,y)

.

Now since f is entire, we have that u and v are di↵erentiable with continuous derivatiesand by the Cauchy-Riemann conditions,

u1(x, y) = v2(x, y),

u2(x, y) = �v1(x, y).

Consider now the partial derivatives of w and q. We have,

q1(x, y) = u1(x,�y) = v2(x,�y),q2(x, y) = �u2(x,�y) = v1(x,�y),w1(x, y) = �v1(x,�y) = u2(x,�y),w2(x, y) = v2(x,�y) = u1(x,�y).

Thus,

q1(x, y) = w2(x, y)

q2(x, y) = w1(x, y).

So, q and w satisfy the Cauchy-Riemann conditions. They also have continuous deriva-tives, so it follows that g is entire.

2. Compute sin(i) and cos(i).

Solution. Recall

sin(z) =eiz � e�iz

2i, cos(z) =

eiz + e�iz

2,

We have,

sin(i) =e�1 � e

2i

cos(i) =e�1 + e

2.

3. Find all values of log(i) and log(ei).

Solution. Recall,

log(z) = ln |z|+ i arg(z).

Thus,

log(i) = ln |i||{z}=0

+i arg(i)

= i⇣⇡2+ k 2⇡

⌘, for k 2 Z,

=

⇢. . . ,�i7⇡

2,�i3⇡

2, i⇡

2, i5⇡

2, i9⇡

2. . .

�.

Similarly,

log(ei) = ln |ei|| {z }=1

+i arg(i)

= 1 + i⇣⇡2+ k 2⇡

⌘, for k 2 Z,

=

⇢. . . , 1� i7⇡

2, 1� i3⇡

2, 1 + i

⇡

2, 1 + i

5⇡

2, 1 + i

9⇡

2. . .

�.

4. Let Log(z) denote the principal branch of log(z) as defined in class (or the textbook).Recall that Log is analytic in the domain D = C\I, where I is the set of non-positive(real) numbers, i.e. I = {z 2 C : Re(z) 0, Im(z) = 0}. Also, let f be a functiondefined as,

f(z1, z2) = Log(z1 z2)� Log(z1)� Log(z2),

where z1 2 C, z2 2 C.

(a) Compute f(z1, z2) when Re(z1) > 0, Re(z2) > 0.

(b) Find z1, z2 in D such that f(z1, z2) 6= 0.

Solution. (a) When Re(z) > 0, we have�⇡/2 < Arg(z) < ⇡/2. Therefore, if Re(z1) > 0,Re(z2) > 0, we have that �⇡ < Arg(z1) + Arg(z2) < ⇡. It follows that if Re(z1) > 0,Re(z2) > 0, then Arg(z1 z2) = Arg(z1) + Arg(z2) (why?). Thus,

Log(z1 z2) = ln(|z1| |z2|) + i Arg(z1, z2)

=

⇣ln |z1|+ iArg(z1)

⌘+

⇣ln |z2|+ i Arg(z1, z2)

⌘

= Log(z1) + Log(z2).

Therefore, f = 0.

(b) Let z1 = z2 = ei3⇡/4. Then, Arg(z1) = 3⇡/4. But Arg(z1 z2) = �⇡/2. Thus,

Log(z1 z2) = �i⇡/2,Log(z1) = Log(z2) = i3⇡/4,

so that f(z1, z2) = �i2⇡.

5. We know that if x is real valued and positive, then log(x2) = 2 log(x). Does the equalityLog(z2) = 2Log(z) hold in the domain D defined above in Q4?

Solution. No. Take z = ei3⇡/4 as above. Log(z2) = �i⇡/2 6= i3⇡/2 = 2Log(z).

6. Determine the domain D0 where the function f(z) = Log(i� z) is analytic.

2


Due 12.11.2014

1. Suppose f(z) is continous in a domain that contains z0. Show that

lim�z!0

Z 1

0f(z0 + t�z)dt = f(z0).

Solution. Suppose ✏ > 0 is given. We want to show that for some � > 0, if |�z| < �,��Z 1

0f(z0 + t�z)dt� f(z0)

�� =��Z 1

0f(z0 + t�z)� f(z0) dt

�� ✏.

Now since f is continuous at z0 we can find a small number " > 0 such that if |u| ", then

|f(z0 + u)� f(z0)| ✏.

Set � = ". Note that if 0 t 1 and |�z| �, then |t�z| �. Thus,��Z 1

0f(z0 + t�z)� f(z0) dt

�� Z 1

0|f(z0 + t�z)� f(z0)| dt

Z 1

0✏ dt

✏.

2. Let � be a positively oriented circle of radius r around the point z0. ComputeZ

�

1

z � z0dz.

Solution. We can either refer to the Cauchy integral formula or compute the integral explicitly, using aparameterization. For the sake of demonstrating the procedure let us do the latter.

We first need a parameterization for �. Note that

z(t) = z0 + r eit, for 0 t 2⇡

is an admissable parameterization. Note also that z0(t) = ir eit. We now compute

Z

�

1

z � z0dz =

Z 2⇡

0

1

z(t)� z0z0(t)dt

=

Z 2⇡

0

1

r eitireitdt

= 2⇡i.

Observe that the result is independent of r.

3. Consider the two circular contours �0 and �1 below.

i

�i

�0

i

�i

�1

(a) Compute

Z

�0

1

zdz.

(b) Make use of the result of part (a) to compute

Z

�1

1

zdz.

Note that you don’t need to parameterize �1 in this case.

Solution. (a) A parameterization for �0 is,

z(t) = eit, for ⇡/2 t 3⇡/2.

Note that z0(t) = ieit. So,

Z

�0

1

zdz =

Z 3⇡/2

⇡/2

1

eitieitdt = ⇡i.

(b) From Q2, we know that

Z

�0��1

1

zdz = 2⇡i.

So,

Z

�1

1

zdz = �⇡i.

4. Compute the integral

Z

C

z + 2

(z2 � 1) dz

where C is the circle of radius two around the origin, traversed in the clockwise direction.

Solution. Let us use the Cauchy integral formula for this question. Notice that

z + 2

z2 � 1 =3

2

1

z � 1 �1

2

1

z + 1.

Thus,

Z

C

z + 2

(z2 � 1) dz =Z

C

3

2

1

z � 1dz +Z

C

3

2

1

z � 1 dz

= 2⇡i

✓3

2� 1

2

◆.

5. (a) Suppose g(x, y) is a real-valued and continuous function. Show that if

h(z) = eig(x,y),

is analytic in a domain D, then the partial derivatives of g(x, y) are zero (so that g is constant).

(b) Suppose f(z) is analytic in a domain D. Show that if |f(z)| is constant, then f(z) is also constant.

Solution. (a) Note that

h(x, y) = cos�g(x, y)

�| {z }

u(x,y)

+i sin�g(x, y)

�| {z }

v(x,y)

.

The Cauchy-Riemann conditions for h are,

ux = � sin(g) gx = cos(g) gy = vyuy = � sin(g) gy = � cos(g) gx = �vx.

Multiplying these two equations, we obtain,

sin2�g(x, y)

�gx(x, y) gy(x, y) = cos

2�g(x, y)

�gx(x, y) gy(x, y).

Since g is continuous, these equations can hold in an open set only if gx(x, y) gy(x, y) = 0. Thus atany (x, y) either gx or gy must be zero. But plugging this into the Cauchy-Riemann equations, we seethat the other partial derivative must also be zero. Thus follows the claim.

2

(b) If |f | is constant (say M), we can express f as f(x, y) = M eig(x,y) for a continuous and real valuedg. Now, by part (a), it follows that g(x, y) is constant, so that f is also constant.

6. (From the textbook) Find all functions f analytic in the unit disk D (i.e. z such that |z| < 1) that alsosatisfy f(0) = i and |f(z)| 1 for all z 2 D.

Solution. It follows by the maximum modulus principle that f(z) = i is the only analytic function thatsatisfies the stated conditions.

7. We noted in class that the series

1X

n=0

cn

converges if |c| < 1. Prove that the series do not converge if |c| � 1.

Solution. Let |c| � 1. Also, let sn denote the partial sums defined as,

sn =nX

k=0

cn.

Assume that the series converges and the limit is u. This means that given any ✏ > 0, we can find Nsuch that if n � N , then |u � sn| ✏. Now suppose ✏ < 1/4 and for some n, |sn � u| ✏. Then,|sn+1�u| = |sn+ cn+1�u| � |cn+1|� |sn�u| � 3/4 > ✏. Thus we cannot find an integer N that satisfiesthe stated condition when ✏ < 1/4. So, the series cannot be convergent.

8. We showed in class that Fn(z) = zn converges uniformly to F (z) = 0 on the set of z with |z| < r, if r < 1.Show that, if r = 1, then Fn(z) do not converge to F (z) uniformly.

Solution. The problem here is that as z approaches 1, convergence slows down. Let us show thisrigorously. Suppose convergence is uniform so that given ✏ > 0, we do find N such that if n � N , then

|Fn(z)| ✏, for all z with |z| < 1.

But now consider u = exp (ln(2✏)/N). Assuming ✏ < 1, we have that u < 1. But u2N = 2✏. Thus,we cannot find an integer N that satisfies the condition above. Actually in this case, convergence is notuniform but pointwise (show pointwise convergence as an exercise).

3


Due 26.11.2014

1. Remember that the principal branch of the logarithm, namely Log(z) is defined everywhere except theorigin. Can you find a Laurent series expansion of Log(z) for C\{0} around the origin?

Solution. No, we cannot find a Laurent series around the origin because we cannot find an annular ringaround the origin that does not intersect the branch cut.

2. Note that the Taylor series of ez is given by

1X

k=0

1

k!zk.

Verify that this series is convergent for all z 2 C by showing that

limn!1

✓1

n!

◆1/n= 0.

3. (From the textbook) Does there exist a power seriesP1

k=0 ak zk which converges at z0 = 1 + 3i but

diverges at z1 = 2 + 2i? If you think so, find one. If not, explain why not.

Solution. Note that |z0| =p10 and |z1| =

p8. But we know that if a Taylor series around the origin

converges for |z| = r, then it converges for all z with |z| < r. Therefore, we cannot find such a series.

4. (a) Suppose f(z) has a Taylor series expansion

f(z) =1X

k=0

ak zk,

uniformly convergent for |z| < R. Find the Taylor series expansion of f 0(z) in terms of ak. What isthe radius of convergence of the series for f 0?

(b) (From the textbook) Two power series of the form

1X

k=0

ak zk and

1X

k=0

k ak zk

have the same radius of convergence. Explain why.

5. Find the Laurent series of cos(1/z) around 0 for |z| > 0.

Solution. Note that the Taylor series for cos(t) is

cos(t) =1X

k=0

(�1)k

(2k)!t2k,

valid for all t in C. Replacing t with 1/z, we obtain the Laurent series around the origin, valid for z 6= 0as,

cos(1/z) =1X

k=0

(�1)k

(2k)!z�2k.

6. Consider the function

f(z) =1

z2 � z .

(a) Find the Laurent series for f(z) around 0 valid for 0 < |z| < 1.(b) Find the Laurent series for f(z) around 0 valid for 1 < |z|.

Solution. Notice that

1

z2 � z =1

z � 1 �1

z.

(a) Note that,

1

z � 1 =1X

k=0

zk,

for |z| < 1. So,

1

z2 � z = �1

z+

1X

k=0

zk,

for 0 < |z| < 1.(b) To obtain a convergent series for |z| > 1, note that

1

z � 1 = �1

z

1

1� z�1 .

Therefore,

1

z � 1 = �1

z

1X

k=0

z�k =1X

k=1

�z�k

for |z�1| < 1, or |z| > 1. So,

1

z2 � z =1X

k=1

�z�k

7. (From the textbook) Suppose f(z) has an isolated singularity at z = 0 but |f(z)| is bounded for 0 < |z| < 1.Show that the singularity is removable.

Solution. Let |f(z)| < M for 0 < |z| < 1 (such an M exists because |f(z)| is given to be boundedin the punctured unit disk). Consider now a Laurent series expansion of f around the origin, valid for0 < |z| < ✏, where the only singularity of f in this region is the origin.

8. Suppose that f(z) is entire and f(z) = 0. Show that f(z)/z is also entire.

9. (From the textbook) Find the smallest r and the greatest R such that the Laurent series

1X

k=�1

zk

2|k|

is convergent if r < |z| < R.

10. What type of a singularity does f(z) = z e1/z have at z = 0?

2


Midterm Examination I

15.10.2014

Student Name :

Student Num. :

4 Questions, 100 Minutes

Please Show Your Work for Full Credit!

1. (a) Consider the function(25 pts)

f(z) = |z|2 + 2|z|+ 1.

Determine whether f is analytic or not. Please briefly explain your answer.

(b) Consider the function (for z = x+ iy with x, y real)

g(z) = x3 � 3x y2 + y + i v(x, y).

Find a real-valued function v(x, y) such that g(z) is analytic.

2. Recall that the principal value of arg(z) is defined for z 6= 0 as,(25 pts)

Arg(z) = ✓ 2 (�⇡,⇡], such that ei✓ = z|z| .

(a) Sketch the largest domain D for which Arg(z) is continuous.

(b) Find a function h(z) so that

f(z) = Arg(z) + i h(z)

is analytic in the domain D̃ = {z : Re(z) > 1}.

(c) Consider another branch of arg(z), defined for z 6= 0 as

A(z) = ✓ 2 (�⇡/4, 7⇡/4], such that ei✓ = z|z| .

Sketch the largest domain D̂ for which A(z) is continuous. Is A(z) also analytic on D̂?

(Please briefly explain your reasoning for credit.)

3. (a) Write down all values of f(z) = z1/4 for z = 8p2(1 + i). How many distinct values can you(25 pts)

find?

(b) Write down all values of g(z) = zi/2 for z = (1 + i)/p2. How many distinct values can you

find?

4. Consider the directed smooth curve � shown below which is a half circle of radius 2, starting at(25 pts)

�0 = 2i and terminating at �1 = �2i.

�2

2i

�2i

�

(a) Find a parameterization for �. That is, find a function z(t) such that all of the following

hold.

• z(0) = �0 = 2i, and z(1) = �1 = �2i.• z(t) is a point of � for 0 t 1.• z0(t) 6= 0 for 0 t 1.

(b) Compute

Z

�

1

z2dz.


Midterm Examination II

26.11.2014

Student Name :

Student Num. :



1. Let � and � be two directed contours (both starting at z0 = 1 and ending at z1 = i) as shown(20 pts)

below.

i

1

�

i

1

�

(a) Compute

Z

�z2 dz.

(b) Compute

Z

�|z|2 dz.

2. Let C be the positively oriented circle of radius 2 around the origin as shown below.(25 pts)

2

C

�2

(a) Compute

Z

C

cos(z2)

z2 � 3z dz.

(b) Suppose g(z) is defined for |z| < 2 as,

g(z) =

Z

C

u2 � 2u� 1(u� 3)(u� z) du.

Find g(0), g(1), g0(0), g0(1).

3. Consider the function(20 pts)

f(z) =1

z2 � z

(a) Find the Laurent series of f(z) around the point z0 = 0, valid for |z| > 1.

(b) Find the Laurent series of f(z) around the point z1 = 1, valid for |z � z1| < 1.

4. Suppose f(z) has a Taylor series expansion of the form(20 pts)

f(z) =1X

k=0

ak zk,

and the series is known to converge for |z| < R, for some R > 0. Let g(z) be defined as,

g(z) =

✓1 +

1

z

◆f 0(z).

(a) Find a Laurent series for g(z) around the point z0 = 0. Express the Laurent series coe�cients

in terms of ak.

(b) What is the region of convergence for the series you found in part (a)? Please briefly explain

your answer for full credit.

5. Let the functions f(z) and g(z) be defined as(15 pts)

f(z) = sin(z)� 1, and g(z) = 1f(z)

.

Notice that z0 = ⇡/2 is an isolated singularity of g(z). Please provide your reasoning while

answering the following questions.

(a) Is z0 a removable singularity for g(z)?

(b) Is z0 a pole of order m for g(z)? If so, what is m?

(c) Is z0 an essential singularity for g(z)?


Final Examination

25.12.2014



1. For z = x+ iy with x, y real, determine whether the functions below are analytic or not. If they(20 pts)

are analytic, compute their derivative at z0 = ⇡ + i⇡ (that is, evaluate f 0(z0)).

(a) f(x, y) = x2 + y2 + i 2xy.

(b) f(x, y) = e�y⇥cos(x) + i sin(x)

⇤.

2. Find all z 2 C that satisfy the following equations.(20 pts)

(a) ez = 1.

(b) z8 = 1.

(c) Log(z) = i⇡/2, where Log(z) denotes the principal value of log(z).

3. Let � and � be two directed contours (both starting at z0 = 1 and terminating at z1 = �1) as(20 pts)shown below.

i

1

�

�1

i

1

�

�1

(a) ComputeZ

�z�2 dz.

(b) ComputeZ

�|z|�2 dz.

4. Let C be the positively oriented circle of radius 4 around the origin as shown below.(20 pts)

4

C

�4

EvaluateZ

C

cos(z2)

z2 � 3z dz.

5. Evaluate the Cauchy principal value of the real valued integral(20 pts)Z 1

�1

1

x2 + 2dx.

mat 205e – theory of complex functionsmat 205e – homework 2 due 01.10.2014 1. consider a...

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