Lecture 1
The 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 introduce
topological 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 (considered
as vectors in R2 ) is the real part of z w̄. Use this to give another proof that multiplication
by a fixed complex number is a conformal linear transformation.
Solution. If z = x + yi and w = p + qi then
z w̄ = (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 let
w and w0 be complex numbers. By what we just remarked, the angle between them is
Re w0 w̄
.
cos−1
|w||w0 |
The angle between zw and zw0 is
cos−1
Re w0 z z̄ w̄
|zw||zw0 |
.
Since z z̄ = |z|2 the factors of z cancel and the angles are equal.
1
Lecture 2
Holomorphic 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 Möbius transformations map circlines to circlines.
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 + y 2 ) + 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 radius
r. If 0 < p < r define q = r2 /p and λ = p/r. The equation |z − p| = λ|z − q then
becomes
p2
r4
2r2
x2 + y 2 − 2px + p2 = 2 x2 + y 2 −
x+ 2 .
r
p
p
After some algebra this gives
x2 + y 2 = r2 ,
the equation of the original circle.
Notice that for a general Mobius transformation M (z) = (az + b)/(cz + d)
az + b a z − p = ,
cz + d c z − q
where p = −b/a and q = −c/d. Thus the equation |M (z)| = constant represents
a circline, and every circline can be so represented. Since the composite of Mobius
transformations is Mobius, it follows that Mobius transformations map circlines to
circlines.
2
Lecture 3
Some examples of holomorphic functions
Exercise 3.1. Show that the zeta function can also be defined by the integral
Z ∞ s−1
t
1
dt,
ζ(s) =
Γ(s) 0 et − 1
provided that Re s > 1.
Solution. For t > 0 we can write
∞
X
e−t
1
=
=
e−nt .
et − 1
1 − e−t
n=1
From the dominated convergence theorem, then,
Z
0
∞
∞ Z ∞
X
ts−1
dt
=
ts−1 e−nt dt.
et − 1
n=1 0
But the substitution u = nt shows that
Z ∞
Z
ts−1 e−nt dt = n−s
0
∞
us−1 e−u du = n−s Γ(s)
0
and so
Z
0
∞
∞
X
ts−1
dt
=
n−s Γ(s) = Γ(s)ζ(s),
et − 1
n=1
as required.
3
Lecture 4
Plane 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 a
holomorphic function g with eg(w) = w, but that there is no function defined on all of
C \ {0} that has this property.
Solution. Let w0 = r0 eiθ0 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 bijectively
onto 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 function
defined on U .
If a holomorphic logarithm function existed on the whole of C \ {0}, it would in
particular give a continuous map from T to iR (the imaginary axis) whose composite
with 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 5
Path integrals and Cauchy’s theorem
Exercise 5.1. Show that Cauchy’s theorem for a triangle remains true if f is continuous
in Ω and differentiable everywhere except at a single point. (Approximate a triangle
with 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 a
triangle ABC. Since f is continuous at A, given > 0 we can find points B 0 (on AB)
and C 0 (on AC) sufficiently close to A that
Z
< .
f
(z)dz
∂AB 0 C 0
Now
Z
Z
f (z)dz =
∂ABC
Z
f (z)dz +
∂AB 0 C 0
Z
f (z)dz +
∂BB 0 C
f (z)dz
∂CB 0 C 0
and the last two integrals vanish by Cauchy’s theorem (original version). Thus
Z
f (z)dz < ∂ABC
and letting → 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 6
Consequences of Cauchy’s theorem
Exercise 6.1. (The reflection principle) Let f be continuous on the closed upper halfplane, 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 (z̄)
when z is in the lower half-plane, then the extended function is holomorphic on the
entire complex plane.
Solution. We
R shall use Morera’s theorem. If T is a triangle in the open upper halfplane, then ∂T f (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 upper
half-plane can be regarded as the limit of a sequence of triangles in the open upper
half-plane.
If T is a triangle in the open lower half-plane then
Z
Z
f (z)dz =
∂T
Z
f (z̄)dz =
∂T
f (w)dw = 0
∂T
using the substitution w = z̄. Again, this extends by continuity to triangles in the
closed lower half-plane.
Any triangle can be subdivided into a sum of triangles in the upper and lower half
planes. Thus the integral of f around any triangle is zero. Morera’s theorem now
implies that f is holomorphic.
6
Lecture 7
The rigidity of holomorphic functions
Exercise 7.1. Show that if f is an entire function and |f (z)| 6 C(|z|n + 1) for some
constant 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) =
∞
X
ak z k
k=0
convergent in C. Let p(z) be the polynomial
p(z) =
n
X
ak z k .
k=0
Then (f (z) − p(z))/z n is a bounded holomorphic function that vanishes at 0. By
Liouville’s theorem, it is identically zero, and so f ≡ p.
7
Lecture 8
The Global Cauchy Theorem
Exercise 8.1. Show that every 1-cycle is equivalent to a formal linear combination of
closed paths.
P
Solution. Let Γ be a 1-cycle, and let it be represented as j kj [γj ]. We can replace it
by an equivalent cycle where all the numbers kj are positive; to do this, notice that if
γ : [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 the
kj . We’ll prove by induction of |Γ| that every 1-cycle is equivalent to a combination
of closed paths. This is apparent if |Γ| = 1 since then Γ = [γ] and the cycle condition
implies that the beginning and ending
P 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 0 . Thus, starting
at 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 there
is 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 the
start of γ1 .
But then Γ0 = [γ1 ] + · · · + [γp ] is a cycle, and it is equivalent to the closed path
obtained by concatenating γ1 , . . . , γp . Moreover, Γ − Γ0 is a cycle, and its height
|Γ − Γ0 | = |Γ| − p < γ. By induction, Γ − Γ0 is equivalent to a sum of closed paths;
so Γ = (Γ − Γ0 ) + Γ0 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, 1}
that is homologous to zero but is not homotopic to zero (i.e., cannot be continuously
deformed to a constant path).
8
Lecture 9
Laurent series and the residue theorem
Exercise 9.1. Suppose that f1 and f2 are as above, possibly having zeroes or poles at
a. What, if anything, can you say about the order of the zero or pole of f1 + f2 at a, in
terms 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 max{N1 , N2 }. However,
if N1 = N2 , then the sum f1 + f2 may have a pole of order N1 or of any lesser
order (including no pole at all). These statements are easily verified by adding Laurent
zeries.
9
Lecture 10
Counting zeroes and poles
10
Lecture 11
Calculations with the residue theorem
Exercise 11.1. Derive the formula for the residue at a double pole: if g has a double
zero at a, then the residue of f /g at a is
0
f (a)g 000 (a)
f (a)
.
−
2
g 00 (a)
3g 00 (a)2
Verify that this is consistent with our solution to Example ??.
Solution. Write
g(z) = (z − a)2 h(z)
where h is holomorphic and h(a) 6= 0. Then (either by direct calculation using the
product rule for differentiation, or more simply by comparing Taylor series on the left
and right hand sides of the above identity) we find
h(a) = 21 g 00 (a),
Now write
h0 (a) = 61 g 000 (a).
f (z)
f (z)/h(z)
=
.
g(z)
(z − a)2
The numerator is holomorphic near a, so the residue is
f 0 (a) f (a)h0 (a)
2f 0 (a) 2f (a)g 000 (a)
d f (z)
=
−
= 00
−
2
dz h(z) z=a
h(a)
h(a)
g (a)
3g 00 (a)2
as required.
11
Lecture 12
The Gamma Function
12
Lecture 13
More examples of contour integration
13
Lecture 14
More about the Riemann Sphere
Exercise 14.1. Let S be a compact connected Riemann surface. Show that the only
holomorphic functions S → C are the constant functions.
Solution. Let f : S → C be holomorphic. Since S is compact, |f | attains a maximum
at some point a ∈ S. By the maximum principle (applied to f ◦ φ−1 , where φ is a chart
near a), f must be constant (say equal to c) on some neighborhood of a. Consider now
the set of all points x ∈ S such that f = c on a neighborhood of x. By definition this
set is open. By the principle of isolated zeroes, it is closed. It is nonempty as we have
seen, and thus (by connectedness) it is the whole of S.
14
Lecture 15
Automorphisms
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 the
group of translations acts transitively on C. Since translations also give automorphisms
of S, we see that any two finite points of S are equivalent under the automorphism
group. The transformation z 7→ 1/z is also an automorphism of S and maps ∞ to a
finite point, so in fact all points of S are equivalent.
15
Lecture 16
Hyperbolic Geometry
Exercise 16.1. Prove the hyperbolic sine law: in a hyperbolic triangle as discussed
above, one has
sinh BC
sinh CA
sinh AB
=
=
.
sin α
sin β
sin γ
Solution. Consider first the special case of a triangle with a right angle at C. Write
d(A, B) = c, d(B, C) = a, d(C, A) = b.
We have
cosh c = cosh a cosh b
cosh b = cosh a cosh c − sinh a sinh c cos β
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 b
=
sinh2 c
sinh2 c
so sin β = sinh b/ sinh c. This is the sine rule for a right triangle. The general case
follows 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 17
The Riemann mapping theorem
17
Lecture 18
Multi-valued functions
Exercise 18.1. Find a region on which a branch of the inverse cosine function is defined. (Start by proving that
p
cos−1 (z) = i log z + z 2 − 1
and then look for a branch of the function on the right.)
Solution. If z = cos w = 12 (eiw + e−iw ), then e−iw is a root of the quadratic equation
t2 − 2zt + 1 = 0. The roots of this equation are
p
t = z ± z 2 − 1,
√
which gives the formula w = i log z + z 2 − 1 from the question. There are branch
√
√
√
points of z 2 − 1 at ±1. Since (z + z 2 − 1)(z − z 2 − 1) = 1, the expression
under the logarithm sign is never zero, so no further branch points are introduced by
the logarithm term. The most natural way to cut the plane here is to cut from −∞ to
−1 and 1 to ∞. The
cuts is a simply connected region, on
√ complement of these two √
which branches of z 2 − 1 and then i log z + z 2 − 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 19
Constructing conformal maps
19
Lecture 20
Contour integrals with multi-valued functions
Exercise 20.1. Evaluate the integral of homework 6, problem 1,
√
Z ∞
x
dx,
2 + 5x + 6
x
0
by means of a contour integral involving a suitably defined branch of the square root
function.
Solution. Let I denote the desired integral.
√ Cut the plane along the positive real axis
and let f (z) be the branch of the function z/(z 2 + 5z + 6) that is positive real-valued
√
just above
√ the real axis. The function f has poles at −2 and −3, with residues i 2
and −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 integrals
along the inner and outer loops tend to 0. Thus, we obtain
√
√
2I = 2πi.(i 2 − i 3),
√
√
so I = π( 3 − 2).
20
Lecture 21
Analytic 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) is
open, so w is continuous.
21
Lecture 22
The monodromy theorem
22
Lecture 23
Basics 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 k∞ =
sup{|f (x)| : x ∈ X \ N }. Indeed, by definition, for every k S
there is a null set Nk such
that |f k∞ = sup{|f (x)| : x ∈ X \ Nk } + 1/k. Take N = Nk , which is a null set.
In particular, this shows that if kf k∞ = 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 k∞ = sup{|fi (x)| : x ∈ X \ Ni },
i = 1, 2.
Let N = N1 ∪ N2 and f = f1 + f2 . Then
kf k∞ 6 sup{|f1 (x)| + |f2 (x)| : x ∈ X \ N }
6 sup{|f1 (x)| : x ∈ X \ N } + sup{|f2 (x)| : x ∈ X \ N } 6 kf1 k∞ + kf2 k∞
23
Lecture 24
The Spaces Lp (X, µ)
24
Lecture 25
The 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 kφkkxk, we see that ex is a bounded linear functional on
E ∗ , with norm kex k 6 kxk. Thus ex ∈ E ∗ ∗, and x 7→ ex is a linear map from E to
E ∗ ∗.
Let us show that this linear map is an isometry (and so, in particular, that it is
injective.) Let x ∈ E. By the Hahn-Banach theorem there is a linear functional φ ∈
E ∗ , of norm 1, such that |φ(x)| = kxk (define φ first on the one dimensional subspace
spanned by x, and extend by Hahn-Banach). Then |ex (φ)| = kxk so kex k > kxk.
We proved the opposite inequality above, so kex k = kxk and x 7→ ex is an isometric
injection.
25
Lecture 26
Applications of the Hahn-Banach Theorem
26
Lecture 27
Convexity and the Hahn-Banach Theorem
Exercise 27.1. Give an example of an absorbing subset in a normed vector space which
does not contain any neighborhood of the origin.
Solution. There are many possible solutions depending on how sophisticated you want
to 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 the
supremum norm, and let A = {f : |f 0 (0)| < 1}. Then A is absorbing, but it
contains no neighborhood of 0 (because one can have very small functions with
very 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 complete
space which is not a 0-neighborhood? The answer is yes. Indeed, if φ : E → R is
a discontinuous linear functional, then {x ∈ E : |φ(x)| < 1} is an absorbing set
that cannot contain a neighborhood of the origin. To produce such a linear functional
on 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 (0) for f ∈ F .
Extend φ to E using the technique of the Hahn-Banach theorem (but without bothering
about the norms at all).
27
Lecture 28
Weak topologies
Exercise 28.1. The argument above, as stated, works for real vector spaces only. Work
out 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 will
produce 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 form
of the Hahn-Banach theorem). Now define Fb to be the inverse image under ψ of the
closed half-plane {z ∈ C : Re(z) 6 c} and complete the proof as before.
28
Lecture 29
Applications of the Baire category theorem
Exercise 29.1. Give an example of a separately continuous map that is not jointly
continuous.
Solution. The function (from R2 to R)
(
(x, y) 7→ f (x, y) =
xy
x2 +y 2
0
(x, y) 6= (0, 0)
(x, y) = (0, 0)
is separately continuous. Indeed, if y 6= 0, f (x, y) is clearly a continuous function
of 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
(, ), for which f (x, y) = 12 , whereas f (0, 0) = 0.
29
Lecture 30
The open mapping and closed graph theorems
Exercise 30.1. Prove the above statements.
Solution. If g ∈ L2 [0, 1] is supported in the interval [, 1] for some > 0, then the
function f (x) = g(x)/x is well-defined and belongs to L2 with kf k 6 −1 kgk. Thus
g 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 T f = 1 then
f (x) = 1/x almost everywhere, and the function 1/x is not in L2 [0, 1]. Thus the range
of T , being dense, cannot be closed (otherwise it would be all of L2 [0, 1]).
30
Lecture 31
Operators on Hilbert Space
Exercise 31.1. Show that in the above situation the norm of Mf is exactly equal to the
L∞ norm of f .
Solution. We need to show that for any > 0 there is g ∈ L2 such that
kMf gk2 > (C − )kf k∞ kgk2 .
(The subscripts denote the norms in L2 and L∞ .) By definition of the L∞ norm, there
is a subset E ⊆ X of positive measure such that |f (x)| > C − for all x ∈ E. Since
we are working with a σ-finite measure space, E contains a subset of positive finite
measure; replace E by this subset and we may assume that E itself has finite measure.
Let g = χE . Then g belongs to L2 with kgk2 = µ(E)1/2 and
Z
kMf gk2 =
1/2
|f |2 dµ
> (C − )µ(E)1/2
E
as required.
31
Lecture 32
More about operators
Exercise 32.1. Use the closed graph theorem to show that if T is a linear map from a
Hilbert space to itself, satisfying
hT u, vi = hu, T vi
for all u, v ∈ H, then T is an operator (i.e., it is bounded).
Solution. Suppose that un → u and T un → w. Then
hT u, vi = hu, T vi = limhun , T vi = limhT un , vi = hw, vi
for every v ∈ H. Thus hT u − w, vi = 0 for every v, whence T u = w. This proves
that the graph of T is closed, so T is continuous by the closed graph theorem. This is
a very early theorem (Hellinger-Toeplitz, around 1903).
32
Lecture 33
Calculations with compact operators
Exercise 33.1. Show that an operator T is compact if and only if it transforms weakly
convergent sequences to norm convergent ones. Use this to get another proof that if the
multiplication operator M in the previous example is compact, then the sequence {an }
tends to zero.
33
Lecture 34
The spectral theorem for compact operators
Exercise 34.1. Extend the two lemmas to normal operators T (those for which T
commutes with T ∗ ).
Solution. Let T be a normal operator. Then
kT xk2 = hT ∗ T x, xi = hT T ∗ x, xi = kT ∗ xk2
and thus ker(T ) = ker(T ∗ ). Applying this to the normal operator T − λI instead of
T gives ker(T − λI) = ker(T ∗ − λ̄I). In other words, every eigenvector T (with
eigenvalue λ) is also an eigenvector for T ∗ (with eigenvalue λ̄).
Now let E be the eigenspace ker(T − λI). If x ∈ E ⊥ and y ∈ E then
hT x, yi = hx, T ∗ yi = hx, λ̄yi = 0.
Thus T x ∈ E ⊥ also.
Let x1 , x2 be eigenvectors for T with eigenvalues λ1 , λ2 . Then
λ2 hx1 , x2 i = hx1 , T x2 i = hT ∗ x1 , x2 i = hλ̄1 , x2 i = λ1 hx1 , x2 i
so either λ1 = λ2 or hx1 , x2 i = 0.
34
Lecture 35
General spectral theory
Exercise 35.1. Prove the spectral mapping theorem for general polynomials p. (The
method 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−1 {µ}.
Arguing as in the special case we see that p(T ) − µI is invertible if and only if all
the T − λk I are invertible. That is, µ belongs to the spectrum of p(T ) if and only if
one of the λk belongs to the spectrum of T . But this is exactly to say that µ belongs to
p(σ(T )).
35
Lecture 36
The spectral radius
36
Lecture 37
Harmonic functions
Exercise 37.1. Show that the assumption of simple connectedness is necessary in the
above theorem. (Consider the function g(x + iy) = log(x2 + y 2 ) on C \ {0}. )
Solution. One can show that g is harmonic by direct calculation, or by observing that
it is locally the real part of a suitably defined branch of log z 2 . But if g were globally
the real part of a holomorphic function f we would have
∂g
∂g
df
=
−i
dz
∂x
∂y
H
by the Cauchy-Riemann equations. Now evaluate f 0 (z)dz around the unit circle
z = cos θ + i sin θ. We obtain
I
Z 2π
f 0 (z)dz =
2(cos θ − i sin θ) · (− sin θ + i cos θ)dθ = 4πi.
0
This contradicts the fundamental theorem of calculus, so no such function f can exist.
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Lecture 38
The mean value property
38
Lecture 39
The Perron construction
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