Complex Analysis
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1 Basics of holomorphic functions
The basic objects of complex analysis are the complex manifold and functions between them. The simplest case is when the manifolds are \(1\)dimensional, in which case they are
called Riemann surfaces. The first important observation (known to Gauss) is that for a holomorphic differential \(1\)form, integrating along a curve only depends on the homology class of the curve. This is known as Cauchy’s
theorem.

Proof. If two curves \(\gamma ,\gamma '\) are homologous via a surface \(M\), then by Stokes’s theorem \(\int _\gamma fdz  \int _{\gamma '}fdz = \int _M dfdz =
0\), as \(f\) is holomorphic, so \(\frac {df}{d\bar {z}} = 0\) and hence \(dfdz =0\). □
Now given a holomorphic differential form \(\omega \) on a Riemann surface \(C\) and a point \(p\), we can think of points \(\gamma \) in the universal cover \(\tilde {C}\) as homotopy classes of paths starting at \(p\). Then
\(\gamma \to \int _\gamma \omega \) gives a function on \(\tilde {C}\) called the primitive of \(\omega \).
For example, if \(\omega \) is \(\frac {1}{z}dz\), then the resulting function is \(\log (z)\). For any simply connected region of the universal cover of \(\CC ^\times \), this descends to a function on an open set of \(\CC
^\times \).
Cauchy’s theorem let’s us compute integrals of meromorphic differential forms in terms of residues. The residue of a differential form \(fdz\) at a point \(p\) is \(a_{1}\) where \(f\) locally is given by \(\sum
_{m}^{1}a_k(zp)^k + H\), where \(H\) is holomorphic. Amazingly enough, this gives an analytic proof of the algebraic fact that \(a_{1}\) is an invariant of the coordinate system.

Theorem 1.3 (Residue formula). If \(fdz\) is a
meromorphic differential form, and \(\gamma \) is a counterclockwise oriented curve bounding a region \(X\), then \(\int _\gamma fdz = 2 \pi i\sum _{p \in X} \res _p(fdz)\).

Proof. By Cauchy’s theorem, this reduces to a local computation around each pole. Then note that the integral for \(\frac {1}{z^k}dz\) is \(0\) for \(k>1\) and \(2 \pi i\) for
\(k = 1\), and since locally the form differs from a holomorphic form by a finite linear combination of these, we are done. □

Theorem 1.4 (Cauchy integral formulas). A holomorphic
function is analytic, and its \(n^{th}\) derivative at a point \(p\) is given by \(n!\int \frac {f(z)}{2 \pi i(zp)^{n+1}}dz\) where the integral is over a small circle around \(p\). The radius of convergence for the power
series for \(f\) at a point is the largest it could possibly be.

Proof. The case \(n =0\) follows from the residue formula, and by differentiating inside the integral, we get the formula for all \(n\). To see analyticity, suppose that \(f\) is analytic in
the neighborhood of \(p\) of radius \(<r\), then let \(z\) be inside this circle. Now integrating over the circle, we compute
\(\seteqnumber{0}{}{0}\)
\begin{align*}
f(z) &= \frac {1}{2 \pi i}\int \frac {f(\xi )}{\xi z} d\xi \\ &= \frac {1}{2 \pi i}\int \frac {f(\xi )}{\xi p} \frac {1}{1\frac {zp}{\xi p}} d\xi \\ &= \frac {1}{2 \pi i}\int
\sum _{n\geq 0}\frac {f(\xi )}{(\xi p)^{n+1}} (zp)^n d\xi \\ &= \frac {1}{2 \pi i}\sum _{n\geq 0}\int \frac {f(\xi )}{(\xi p)^{n+1}} d\xi (zp)^n \\ &= \sum _{n\geq 0} \frac
{f^{(n)}(p)}{n!}(zp)^n
\end{align*}
□
The integral formulas have an interpretation via Fourier series. Let \(f\) be a holomorphic function on the unit disc.

Proof. Simply define \(f\) at \(p\) via the integral formula, or alternatively, note that we can extend \((zp)^2f\) holomorphically to be \(0\) at \(p\), and then divide by the factor we
put in. □
The unit disk and the complex plane are not isomorphic Riemann surfaces:
Here is a converse to Cauchy’s theorem.

Proof. Note that we can locally define a holomorphic primitive of \(f\) using this fact, and it is analytic, so its derivative, \(f\), is as well. □

Lemma 1.11. If \(f(z,s)\) is continuous and holomorphic in the variable \(z\), then \(\int
_{[0,1]}f(z,s)\) is holomorphic, and its derivatives are \(\int _{[0,1]}f^{(n)}(z,s)\).

Proof. Consider the series expansion of \(f\) around the limit point \(p\), and note that for a small enough disk around \(p\), \(f\) cannot vanish, as it behaves like the first
nonvanishing term. Thus the set of points on which \(f\) is locally identically \(0\) is open, but it is also closed so we are done. □

Corollary 1.13 (Analytic continuation). If \(f,g\) are
defined on regions that intersect on a connected region, and the set of points on which they agree have a limit point, then \(f \equiv g\).

Proof. WLOG we are looking at \(f\) near \(0\) and \(f(0) = 0\). Then \(f\) has a power series expansion, so for some \(k\), \(\frac {f}{z^{k}}\) has a power series expansion
starting with a nonzero constant \(c\). Since \(\frac {f}{z^k}\) is continuous, for a small enough neighborhood around \(0\), \(\frac {f}{z^k}\) is the \(k^{th}\) power of a holomorphic function \(g\) since the map
\(z\mapsto z^k\) is a covering map away from \(0\). Thus \(f = (gz)^k\) locally, and \(gz\) locally has an inverse. □

Proof. A nonconstant polynomial \(p\) is a holomorphic function from \(\PP ^1 \to \PP ^1\) so it has closed image, but it also is an open map, so must be surjective. Alternatively if
there were no roots, \(\frac {1}{p}\) would be an entire bounded holomorphic function, so would be constant. □

Proof. Note that such a meromorphic function \(f\) has finitely many zeros and poles. after multiplying by a polynomial, we can remove all the poles in \(\CC \). But then we have have
at worst a pole at infinity, so \(f(z)\) grows at most like \(cz^k\) where \(k\) is the order of the pole. But then by the Cauchy inequalities, some derivative of \(f\) is \(0\), so \(f\) is a polynomial. □

Theorem 1.19 (Schwarz reflection principle). Suppose that
\(f\) is holomorphic on a region \(D^+\) in the upper half plane and extends continuously to a realvalued function \(\partial D \cap \RR \). Then if \(D^{}\) is \(D\) reflected along \(\RR \), then \(f\) extends to a
holomorphic function on the interior of \(D^+\cup D^\).

Proof. Define \(f\) on \(D^\) by \(f(\bar {z}) = \overline {f(z)}\), and note that it is holomorphic on the interior, since it is still given by a power series around every point.
Moreover, by Morera’s theorem it is holomorphic at the boundary connecting \(D^+,D^\). □

Theorem 1.20 (Argument principle). For \(f\)
meromorphic, \(\frac {1}{2 \pi i } \int \frac {f'(z)}{f(z)}dz\), integrated over a closed curve bounding a region \(X\) is the number of zeros minus the number of poles of \(f\) in the region, counted with
multiplicity.

Proof. First note that the formula in the statement is a homomorphism from meromorphic functions to \(\RR \). Then note that for a nonvanishing holomorphic function, it is \(0\),
and that for \((zz_0)^k\), where \(z_0\) is in \(X\), it is \(k\). □
The following theorem can be used to give an alternate proof of the open mapping theorem.

Theorem 1.21 (Rouché’s theorem). If,
\(f>g\) are holomorphic on the interior of an open set, then \(f,f+g\) have the same number of zeros.

Theorem 1.22 (Runge’s approximation theorem). If \(f\) is
holomorphic on an open set \(U\) and \(K\) is compact in \(U\), then \(f\) can be uniformly approximated on \(K\) by meromorphic functions, and if \(K^c\) is connected, it can be uniformly approximated by polynomials.

Proof. \(K\) is contained in finitely many squares \(D_i\) that are part of a grid and inside \(U\). \(f(z)\) can be given by \(\frac {1}{2 \pi i}\sum _i \int _{D_i} \frac
{f(\zeta )}{\zeta z}d\zeta \), and each integral can be uniformly approximated by Riemann sums, which are rational functions. Finally, consider the set \(S\) of \(z_0\) such that \(\frac {1}{zz_0}\) can be uniformly
approximated by polynomials. Using the geometric series, since \(K\) is bounded, sufficiently large \(z_0\) are in \(S\). Now clearly \(S\) is closed, and to see \(S\) is open, simply note that \(\frac {1}{zz_1} = \frac
{1}{zz_0}\frac {1}{1\frac {z_1z_0}{zz_0}}\), and so for \(z_1\) close to \(z_0\), the geometric series again gives a uniform approximation. □