Lecture 1The complex plane

Exercise 1.1. Show that the modulus obeys the triangle inequality

|z ± w| 6 |z|+ |w|.

This allows us to make the complex plane into a metric space, and thus to introducetopological notions such as open and closed sets, continuity. etc.

Solution. See Proposition 1.8, last semester.

Exercise 1.2. Show that the dot product of two complex numbers z and w (consideredas vectors in R2) is the real part of zw. Use this to give another proof that multiplicationby a fixed complex number is a conformal linear transformation.

Solution. If z = x+ yi and w = p+ qi then

zw = (x+ yi)(p− qi) = (xp+ yq) + (yp− xq)i.

The real part of this is the dot product of z and w, thought of as vectors in R2. Now letw and w′ be complex numbers. By what we just remarked, the angle between them is

cos−1

(Rew′w|w||w′|

).

The angle between zw and zw′ is

cos−1

(Rew′zzw|zw||zw′|

).

Since zz = |z|2 the factors of z cancel and the angles are equal.

1

Lecture 2Holomorphic Maps

Exercise 2.1. Show that every equation of the form

|z − p| = λ|z − q|

defines a circline, and that every circline can be expressed (not uniquely) in this way.Use this fact to give another proof that Mobius transformations map circlines to cir-clines.

Solution. If you take the equation |z − p|2 = λ2|z − q|2 and expand in terms of (x, y)with z = x+ iy, you get

(1− λ2)(x2 + y2) + linear terms in x, y + constant = 0.

This represents a straight line if λ = 1, and a circle otherwise.If λ = 1, |z− p| = |z− q| represents the perpendicular bisector of the segment pq;

any line can be so represented. For circles, consider wlog the circle center 0 and radiusr. If 0 < p < r define q = r2/p and λ = p/r. The equation |z − p| = λ|z − q thenbecomes

x2 + y2 − 2px+ p2 =p2

r2

(x2 + y2 − 2r2

px+

r4

p2

).

After some algebra this givesx2 + y2 = r2,

the equation of the original circle.Notice that for a general Mobius transformation M(z) = (az + b)/(cz + d)∣∣∣∣az + b

cz + d

∣∣∣∣ =∣∣∣ac

∣∣∣ ∣∣∣∣z − p

z − q

∣∣∣∣ ,where p = −b/a and q = −c/d. Thus the equation |M(z)| = constant representsa circline, and every circline can be so represented. Since the composite of Mobiustransformations is Mobius, it follows that Mobius transformations map circlines tocirclines.

2

Lecture 3Some examples of holomorphic functions

Exercise 3.1. Show that the zeta function can also be defined by the integral

ζ(s) =1

Γ(s)

∫ ∞

0

ts−1

et − 1dt,

provided that Re s > 1.

Solution. For t > 0 we can write

1et − 1

=e−t

1− e−t=

∞∑n=1

e−nt.

From the dominated convergence theorem, then,∫ ∞

0

ts−1

et − 1dt =

∞∑n=1

∫ ∞

0

ts−1e−ntdt.

But the substitution u = nt shows that∫ ∞

0

ts−1e−ntdt = n−s

∫ ∞

0

us−1e−udu = n−sΓ(s)

and so ∫ ∞

0

ts−1

et − 1dt =

∞∑n=1

n−sΓ(s) = Γ(s)ζ(s),

as required.

3

Lecture 4Plane topology and complex analysis

Exercise 4.1. Prove the facts corresponding to those above for the logarithm function.That is, show that every nonzero w0 has a neighborhood on which there is defined aholomorphic function g with eg(w) = w, but that there is no function defined on all ofC \ 0 that has this property.

Solution. Let w0 = r0eiθ0 be nonzero. Define Ω to be the strip x+ iy : x ∈ R, θ0 −

π < yθ0 + π in C. Then the exponential function f(z) = ez maps Ω bijectivelyonto U = reiθ : r > 0, θ0 − π < θ < θ0 + π, which is an open set (a cut plane)containing w0. Lemma ?? now shows that there is a holomorphic logarithm functiondefined on U .

If a holomorphic logarithm function existed on the whole of C \ 0, it would inparticular give a continuous map from T to iR (the imaginary axis) whose compositewith the exponential map iR → T would be the identity. But since iR is contractible,the induced map on fundamental groups

π1(T) → π1(iR) = 0 → π1(T)

would be the zero map, and this is impossible.

4

Lecture 5Path integrals and Cauchy’s theorem

Exercise 5.1. Show that Cauchy’s theorem for a triangle remains true if f is continuousin Ω and differentiable everywhere except at a single point. (Approximate a trianglewith the ‘bad’ point as vertex by nearby triangles that don’t contain the bad point.)

Solution. First of all, consider the case where the ‘bad’ point is the vertex A of atriangle ABC. Since f is continuous at A, given ε > 0 we can find points B′ (on AB)and C ′ (on AC) sufficiently close to A that∣∣∣∣∫

∂AB′C′f(z)dz

∣∣∣∣ < ε.

Now ∫∂ABC

f(z)dz =∫

∂AB′C′f(z)dz +

∫∂BB′C

f(z)dz +∫

∂CB′C′f(z)dz

and the last two integrals vanish by Cauchy’s theorem (original version). Thus∣∣∣∣∫∂ABC

f(z)dz∣∣∣∣ < ε

and letting ε→ 0 we get the result.If the bad point is on an edge or inside the triangle, subdivide into two or three

smaller triangles with the bad point at the vertex and apply the previous result.

5

Lecture 6Consequences of Cauchy’s theorem

Exercise 6.1. (The reflection principle) Let f be continuous on the closed upper half-plane, holomorphic on the open upper half-plane, and real-valued on the real axis.Show that, if we extend f to the whole of C by defining

f(z) = f(z)

when z is in the lower half-plane, then the extended function is holomorphic on theentire complex plane.

Solution. We shall use Morera’s theorem. If T is a triangle in the open upper half-plane, then

∫∂Tf(z)dz = 0 by cauchy’s theorem. This result extends to triangles in

the closed upper half-plane, since f is continuous and any triangle in the closed upperhalf-plane can be regarded as the limit of a sequence of triangles in the open upperhalf-plane.

If T is a triangle in the open lower half-plane then∫∂T

f(z)dz =∫

∂T

f(z)dz =∫

∂T

f(w)dw = 0

using the substitution w = z. Again, this extends by continuity to triangles in theclosed lower half-plane.

Any triangle can be subdivided into a sum of triangles in the upper and lower halfplanes. Thus the integral of f around any triangle is zero. Morera’s theorem nowimplies that f is holomorphic.

6

Lecture 7The rigidity of holomorphic functions

Exercise 7.1. Show that if f is an entire function and |f(z)| 6 C(|z|n + 1) for someconstant C, then f is in fact a polynomial (of degree at most n).

Solution. Since f is entire, it has a Taylor series expansion

f(z) =∞∑

k=0

akzk

convergent in C. Let p(z) be the polynomial

p(z) =n∑

k=0

akzk.

Then (f(z) − p(z))/zn is a bounded holomorphic function that vanishes at 0. ByLiouville’s theorem, it is identically zero, and so f ≡ p.

7

Lecture 8The Global Cauchy Theorem

Exercise 8.1. Show that every 1-cycle is equivalent to a formal linear combination ofclosed paths.

Solution. Let Γ be a 1-cycle, and let it be represented as∑

j kj [γj ]. We can replace itby an equivalent cycle where all the numbers kj are positive; to do this, notice that ifγ : [a, b] → C is a path, then the cycle [γ] is equivalent to −[γ\], where γ\ is defined by

γ\(t) = γ(a+ b− t).

Assume then that all the kj are positive, and define the ‘height’ |Γ| to be the sum of thekj . We’ll prove by induction of |Γ| that every 1-cycle is equivalent to a combinationof closed paths. This is apparent if |Γ| = 1 since then Γ = [γ] and the cycle conditionimplies that the beginning and ending points of γ are the same.

Now given a general cycle Γ =∑kj [γj ] (with all k’s positive), the cycle condition

implies that each end point of one γj is the start point of another γj′ . Thus, startingat γ1, we can find a chain of γ’s each of which begins where the previous one ends.Because there are only finitely many γ’s this chain must close at some point, so thereis a finite sequence of γ’s — by renumbering we may assume that it is γ1, . . . , γp —such that the end of γj is the start of γj+1 for 1 6 j 6 p − 1, and the end of γp is thestart of γ1.

But then Γ′ = [γ1] + · · · + [γp] is a cycle, and it is equivalent to the closed pathobtained by concatenating γ1, . . . , γp. Moreover, Γ − Γ′ is a cycle, and its height|Γ − Γ′| = |Γ| − p < γ. By induction, Γ − Γ′ is equivalent to a sum of closed paths;so Γ = (Γ− Γ′) + Γ′ is equivalent to a sum of closed paths as well.

Exercise 8.2. Give an example of a cycle (in fact a closed path) in Ω = C \ 0, 1that is homologous to zero but is not homotopic to zero (i.e., cannot be continuouslydeformed to a constant path).

8

Lecture 9Laurent series and the residue theorem

Exercise 9.1. Suppose that f1 and f2 are as above, possibly having zeroes or poles ata. What, if anything, can you say about the order of the zero or pole of f1 + f2 at a, interms of the orders of f1 and f2?

Solution. Let f1 and f2 have poles of orders N1 and N2 respectively. If N1 6= N2

then ‘the worst singularity wins’: f1 + f2 has a pole of order maxN1, N2. However,if N1 = N2, then the sum f1 + f2 may have a pole of order N1 or of any lesserorder (including no pole at all). These statements are easily verified by adding Laurentzeries.

9

Lecture 10Counting zeroes and poles

10

Lecture 11Calculations with the residue theorem

Exercise 11.1. Derive the formula for the residue at a double pole: if g has a doublezero at a, then the residue of f/g at a is

2(f ′(a)g′′(a)

− f(a)g′′′(a)3g′′(a)2

).

Verify that this is consistent with our solution to Example ??.

Solution. Writeg(z) = (z − a)2h(z)

where h is holomorphic and h(a) 6= 0. Then (either by direct calculation using theproduct rule for differentiation, or more simply by comparing Taylor series on the leftand right hand sides of the above identity) we find

h(a) = 12g′′(a), h′(a) = 1

6g′′′(a).

Now writef(z)g(z)

=f(z)/h(z)(z − a)2

.

The numerator is holomorphic near a, so the residue is[d

dz

f(z)h(z)

]z=a

=f ′(a)h(a)

− f(a)h′(a)h(a)2

=2f ′(a)g′′(a)

− 2f(a)g′′′(a)3g′′(a)2

as required.

11

Lecture 12The Gamma Function

12

Lecture 13More examples of contour integration

13

Lecture 14More about the Riemann Sphere

Exercise 14.1. Let S be a compact connected Riemann surface. Show that the onlyholomorphic functions S → C are the constant functions.

Solution. Let f : S → C be holomorphic. Since S is compact, |f | attains a maximumat some point a ∈ S. By the maximum principle (applied to f φ−1, where φ is a chartnear a), f must be constant (say equal to c) on some neighborhood of a. Consider nowthe set of all points x ∈ S such that f = c on a neighborhood of x. By definition thisset is open. By the principle of isolated zeroes, it is closed. It is nonempty as we haveseen, and thus (by connectedness) it is the whole of S.

14

Lecture 15Automorphisms

Exercise 15.1. Show that the automorphism groups of S and C are transitive.

Solution. Translations z 7→ z + c are automorphisms of C, and it is obvious that thegroup of translations acts transitively on C. Since translations also give automorphismsof S, we see that any two finite points of S are equivalent under the automorphismgroup. The transformation z 7→ 1/z is also an automorphism of S and maps ∞ to afinite point, so in fact all points of S are equivalent.

15

Lecture 16Hyperbolic Geometry

Exercise 16.1. Prove the hyperbolic sine law: in a hyperbolic triangle as discussedabove, one has

sinhBCsinα

=sinhCA

sinβ=

sinhABsin γ

.

Solution. Consider first the special case of a triangle with a right angle at C. Writed(A,B) = c, d(B,C) = a, d(C,A) = b.

We have

cosh c =cosh a cosh bcosh b =cosh a cosh c− sinh a sinh c cosβ

Rewrite the second equation to give

sin2 β = 1− cos2 β =sinh2 a sinh2 c− (cosh b− cosh a cosh c)2

sinh2 a sinh2 c.

Substitute cosh a = cosh c/ cosh b from the first equation and sinh2 a = cosh2 a − 1(standard identity). After canceling a common factor of cosh2 c− cosh2 b we get

cosh2 b− cosh2 c+ sinh2 c

sinh2 c=

sinh2 b

sinh2 c

so sinβ = sinh b/ sinh c. This is the sine rule for a right triangle. The general casefollows by dropping a perpendicular from a vertex of the triangle to the opposite side,and thus dividing it into two right triangles (exactly as in Euclidean geometry).

16

Lecture 17The Riemann mapping theorem

17

Lecture 18Multi-valued functions

Exercise 18.1. Find a region on which a branch of the inverse cosine function is de-fined. (Start by proving that

cos−1(z) = i log(z +

√z2 − 1

)and then look for a branch of the function on the right.)

Solution. If z = cosw = 12 (eiw + e−iw), then e−iw is a root of the quadratic equation

t2 − 2zt+ 1 = 0. The roots of this equation are

t = z ±√z2 − 1,

which gives the formulaw = i log(z +

√z2 − 1

)from the question. There are branch

points of√z2 − 1 at ±1. Since (z +

√z2 − 1)(z −

√z2 − 1) = 1, the expression

under the logarithm sign is never zero, so no further branch points are introduced bythe logarithm term. The most natural way to cut the plane here is to cut from −∞ to−1 and 1 to ∞. The complement of these two cuts is a simply connected region, onwhich branches of

√z2 − 1 and then i log

(z +

√z2 − 1

)can be defined. Note that it

does not suffice to cut from −1 to 1; the square root is well-defined on this cut region,but the logarithm is not.

18

Lecture 19Constructing conformal maps

19

Lecture 20Contour integrals with multi-valued functions

Exercise 20.1. Evaluate the integral of homework 6, problem 1,∫ ∞

0

√x

x2 + 5x+ 6dx,

by means of a contour integral involving a suitably defined branch of the square rootfunction.

Solution. Let I denote the desired integral. Cut the plane along the positive real axisand let f(z) be the branch of the function

√z/(z2 +5z+6) that is positive real-valued

just above the real axis. The function f has poles at −2 and −3, with residues i√

2and −i

√3 respectively. Integrate around the ‘keyhole’ contour described above. The

integrals along the top and bottom sides of the keyhole approach 2I , and the integralsalong the inner and outer loops tend to 0. Thus, we obtain

2I = 2πi.(i√

2− i√

3),

so I = π(√

3−√

2).

20

Lecture 21Analytic Continuation

Exercise 21.1. Complete the proof that w is continuous.

Solution. With notation as in the previous proof, let Ω ⊆ C be open and let p ∈w−1(Ω). Let the germ p be represented by (U, f, a), with f(a) ∈ Ω. Then f(U ∩f−1(Ω)) is a basic neighborhood of p and is contained in w−1(Ω). Thus w−1(Ω) isopen, so w is continuous.

21

Lecture 22The monodromy theorem

22

Lecture 23Basics of Banach Spaces

Exercise 23.1. Show that L∞ is a normed vector space.

Solution. First let us prove that if f ∈ L∞ then there is a null set such that |f‖∞ =sup|f(x)| : x ∈ X \N. Indeed, by definition, for every k there is a null set Nk suchthat |f‖∞ = sup|f(x)| : x ∈ X \Nk+ 1/k. Take N =

⋃Nk, which is a null set.

In particular, this shows that if ‖f‖∞ = 0, then f = 0 almost everywhere.Now to prove the triangle inequality. Let f1 and f2 belong to L∞. By the above

there are null sets N1 and N2 such that

|fi‖∞ = sup|fi(x)| : x ∈ X \Ni, i = 1, 2.

Let N = N1 ∪N2 and f = f1 + f2. Then

‖f‖∞ 6 sup|f1(x)|+ |f2(x)| : x ∈ X \N6 sup|f1(x)| : x ∈ X \N+ sup|f2(x)| : x ∈ X \N 6 ‖f1‖∞ + ‖f2‖∞

23

Lecture 24The Spaces Lp(X,µ)

24

Lecture 25The Hahn-Banach Theorem

Exercise 25.1. For any Banach space E construct an isometric injection E → E∗∗.

Solution. Any x ∈ E defines a linear map ex : E∗ → C via the equation

ex(φ) = φ(x).

Since |ex(φ)| = |φ(x)| 6 ‖φ‖‖x‖, we see that ex is a bounded linear functional onE∗, with norm ‖ex‖ 6 ‖x‖. Thus ex ∈ E∗∗, and x 7→ ex is a linear map from E toE∗∗.

Let us show that this linear map is an isometry (and so, in particular, that it isinjective.) Let x ∈ E. By the Hahn-Banach theorem there is a linear functional φ ∈E∗, of norm 1, such that |φ(x)| = ‖x‖ (define φ first on the one dimensional subspacespanned by x, and extend by Hahn-Banach). Then |ex(φ)| = ‖x‖ so ‖ex‖ > ‖x‖.We proved the opposite inequality above, so ‖ex‖ = ‖x‖ and x 7→ ex is an isometricinjection.

25

Lecture 26Applications of the Hahn-Banach Theorem

26

Lecture 27Convexity and the Hahn-Banach Theorem

Exercise 27.1. Give an example of an absorbing subset in a normed vector space whichdoes not contain any neighborhood of the origin.

Solution. There are many possible solutions depending on how sophisticated you wantto get. Here are a couple.

• Take C as a 2-dimensional vector space over R, and let A be the set

reiθ : 0 < θ 6 2π, 0 6 r 6 θ.

This is easily seen to be absorbing, but it contains no disk around 0.

• Let E be the vector space of differentiable functions [−1, 1] → R, with thesupremum norm, and let A = f : |f ′(0)| < 1. Then A is absorbing, but itcontains no neighborhood of 0 (because one can have very small functions withvery large derivatives).

Both of these examples feel a bit like cheating: in (a), the set A is not convex and in(b), the space E is not complete. Can one have a convex absorbing set in a completespace which is not a 0-neighborhood? The answer is yes. Indeed, if φ : E → R isa discontinuous linear functional, then x ∈ E : |φ(x)| < 1 is an absorbing setthat cannot contain a neighborhood of the origin. To produce such a linear functionalon a complete space needs some transfinite machinery again. For instance, take E =C[−1, 1], F the subspace of functions differentiable at 0, and φ(f) = f ′(0) for f ∈ F .Extend φ to E using the technique of the Hahn-Banach theorem (but without botheringabout the norms at all).

27

Lecture 28Weak topologies

Exercise 28.1. The argument above, as stated, works for real vector spaces only. Workout how to generalize it to the complex case.

Solution. If E is a complex vector space, and A and b are as above, the argument willproduce a bounded real-linear functional φ such that φ(x) < c < φ(b) for all x ∈ A.This φ is the real part of the bounded complex-linear functional

ψ(x) = φ(x)− iφ(ix)

(this is the same complexification trick that we used in the proof of the complex formof the Hahn-Banach theorem). Now define Fb to be the inverse image under ψ of theclosed half-plane z ∈ C : Re(z) 6 c and complete the proof as before.

28

Lecture 29Applications of the Baire category theorem

Exercise 29.1. Give an example of a separately continuous map that is not jointlycontinuous.

Solution. The function (from R2 to R)

(x, y) 7→ f(x, y) =

xy

x2+y2 (x, y) 6= (0, 0)0 (x, y) = (0, 0)

is separately continuous. Indeed, if y 6= 0, f(x, y) is clearly a continuous functionof x, and if y = 0, f(x, y) = 0 identically. Similarly for x fixed as a function of y.However f is not jointly continuous, because any neighborhood of 0 contains points(ε, ε), for which f(x, y) = 1

2 , whereas f(0, 0) = 0.

29

Lecture 30The open mapping and closed graph theorems

Exercise 30.1. Prove the above statements.

Solution. If g ∈ L2[0, 1] is supported in the interval [ε, 1] for some ε > 0, then thefunction f(x) = g(x)/x is well-defined and belongs to L2 with ‖f‖ 6 ε−1‖g‖. Thusg belongs to the range of T . But now for any g ∈ L2 the functions gn = gχ[1/n,1]

belong to the range of T and gn → g in L2 by the dominated convergence theorem.Thus the range of T is dense.

However, the constant function 1 is not in the range of T , since if Tf = 1 thenf(x) = 1/x almost everywhere, and the function 1/x is not in L2[0, 1]. Thus the rangeof T , being dense, cannot be closed (otherwise it would be all of L2[0, 1]).

30

Lecture 31Operators on Hilbert Space

Exercise 31.1. Show that in the above situation the norm of Mf is exactly equal to theL∞ norm of f .

Solution. We need to show that for any ε > 0 there is g ∈ L2 such that

‖Mfg‖2 > (C − ε)‖f‖∞‖g‖2.

(The subscripts denote the norms in L2 and L∞.) By definition of the L∞ norm, thereis a subset E ⊆ X of positive measure such that |f(x)| > C − ε for all x ∈ E. Sincewe are working with a σ-finite measure space, E contains a subset of positive finitemeasure; replace E by this subset and we may assume that E itself has finite measure.Let g = χE . Then g belongs to L2 with ‖g‖2 = µ(E)1/2 and

‖Mfg‖2 =(∫

E

|f |2dµ)1/2

> (C − ε)µ(E)1/2

as required.

31

Lecture 32More about operators

Exercise 32.1. Use the closed graph theorem to show that if T is a linear map from aHilbert space to itself, satisfying

〈Tu, v〉 = 〈u, Tv〉

for all u, v ∈ H , then T is an operator (i.e., it is bounded).

Solution. Suppose that un → u and Tun → w. Then

〈Tu, v〉 = 〈u, Tv〉 = lim〈un, T v〉 = lim〈Tun, v〉 = 〈w, v〉

for every v ∈ H . Thus 〈Tu − w, v〉 = 0 for every v, whence Tu = w. This provesthat the graph of T is closed, so T is continuous by the closed graph theorem. This isa very early theorem (Hellinger-Toeplitz, around 1903).

32

Lecture 33Calculations with compact operators

Exercise 33.1. Show that an operator T is compact if and only if it transforms weaklyconvergent sequences to norm convergent ones. Use this to get another proof that if themultiplication operatorM in the previous example is compact, then the sequence antends to zero.

33

Lecture 34The spectral theorem for compact operators

Exercise 34.1. Extend the two lemmas to normal operators T (those for which Tcommutes with T ∗).

Solution. Let T be a normal operator. Then

‖Tx‖2 = 〈T ∗Tx, x〉 = 〈TT ∗x, x〉 = ‖T ∗x‖2

and thus ker(T ) = ker(T ∗). Applying this to the normal operator T − λI instead ofT gives ker(T − λI) = ker(T ∗ − λI). In other words, every eigenvector T (witheigenvalue λ) is also an eigenvector for T ∗ (with eigenvalue λ).

Now let E be the eigenspace ker(T − λI). If x ∈ E⊥ and y ∈ E then

〈Tx, y〉 = 〈x, T ∗y〉 = 〈x, λy〉 = 0.

Thus Tx ∈ E⊥ also.Let x1, x2 be eigenvectors for T with eigenvalues λ1, λ2. Then

λ2〈x1, x2〉 = 〈x1, Tx2〉 = 〈T ∗x1, x2〉 = 〈λ1, x2〉 = λ1〈x1, x2〉

so either λ1 = λ2 or 〈x1, x2〉 = 0.

34

Lecture 35General spectral theory

Exercise 35.1. Prove the spectral mapping theorem for general polynomials p. (Themethod is the same.)

Solution. Let p be a polynomial of degree n. By the fundamental theorem of algebra,for each µ there exist λ1, . . . , λn such that

p(z)− µ = c(z − λ1) · · · (z − λn)

where c is a constant. The set λ1, . . . , λn is exactly p−1µ.Arguing as in the special case we see that p(T )− µI is invertible if and only if all

the T − λkI are invertible. That is, µ belongs to the spectrum of p(T ) if and only ifone of the λk belongs to the spectrum of T . But this is exactly to say that µ belongs top(σ(T )).

35

Lecture 36The spectral radius

36

Lecture 37Harmonic functions

Exercise 37.1. Show that the assumption of simple connectedness is necessary in theabove theorem. (Consider the function g(x+ iy) = log(x2 + y2) on C \ 0. )

Solution. One can show that g is harmonic by direct calculation, or by observing thatit is locally the real part of a suitably defined branch of log z2. But if g were globallythe real part of a holomorphic function f we would have

df

dz=∂g

∂x− i

∂g

∂y

by the Cauchy-Riemann equations. Now evaluate∮f ′(z)dz around the unit circle

z = cos θ + i sin θ. We obtain∮f ′(z)dz =

∫ 2π

0

2(cos θ − i sin θ) · (− sin θ + i cos θ)dθ = 4πi.

This contradicts the fundamental theorem of calculus, so no such function f can exist.

37

Lecture 38The mean value property

38

Lecture 39The Perron construction

39