# 1 Useful Techniques

## 1.1 Notation

Notation Definition
$${\mathbb{D}}\mathrel{\vcenter{:}}=\left\{{z {~\mathrel{\Big|}~}{\left\lvert {z} \right\rvert} \leq 1}\right\}$$ The unit disc
$${\mathbb{H}}\mathrel{\vcenter{:}}=\left\{{x+iy {~\mathrel{\Big|}~}y > 0}\right\}$$ The upper half-plane
$$X_{1\over 2}$$ A “half version of $$X$$,” see examples
$${\mathbb{H}}_{1\over 2}$$ The first quadrant
$${\mathbb{D}}_{1\over 2}$$ The portion of the first quadrant inside the unit disc
$$S \mathrel{\vcenter{:}}=\left\{{x + iy {~\mathrel{\Big|}~}x\in {\mathbb{R}},\, 0<y<\pi}\right\}$$ The horizonta strip

If you want to show that a function $$f$$ is constant, try one of the following:

• Write $$f = u + iv$$ and use Cauchy-Riemann to show $$u_x, u_y = 0$$, etc.
• Show that $$f$$ is entire and bounded.

If you additionally want to show $$f$$ is zero, try one of these:

• Show $$f$$ is entire, bounded, and $$\lim_{z\to\infty} f(z) = 0$$.

## 1.2 Greatest Hits

Things to know well:

• Estimates for derivatives, mean value theorem
• Properties of linear fractional transformations
• Automorphisms of $${\mathbb{D}}, {\mathbb{C}}, {\mathbb{CP}}^1$$.

## 1.3 Basic but Useful Facts

• $$z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu = {\left\lvert {z} \right\rvert}^2$$ \begin{align*} \Re(z) = { z + \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \over 2} && && \Im(z) = {z - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \over 2i} .\end{align*}

• $$\operatorname{Arg}(z/w) = \operatorname{Arg}(z) - \operatorname{Arg}(w)$$.

• Exponential forms of cosine and sine: \begin{align*} \cos(\theta) = \frac 1 2 \qty{e^{i\theta} + e^{-i\theta}} && && \sin(\theta) = \frac{1}{2i}\qty{e^{i\theta} - e^{-i\theta}} .\end{align*}

• Various differentials: \begin{align*} dz &= dx + i~dy \\ d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu &= dx - i~dy \\ \\ f_z &= f_x = f_y / i .\end{align*}

• Integral of a complex exponential: \begin{align*} \int_{0}^{2 \pi} e^{i \ell x} d x &=\left\{\begin{array}{ll} {2 \pi} & {\ell=0} \\ {0} & \text{else} \end{array}\right. .\end{align*}

\begin{align*} \sum_{k=1}^{n} k &=\frac{n(n+1)}{2} \\ \sum_{k=1}^{n} k^{2} &=\frac{n(n+1)(2 n+1)}{6} \\ \sum_{k=1}^{n} k^{3} &=\frac{n^{2}(n+1)^{2}}{4} \\ \log(z) &= \sum_{n=0}^\infty { (-1)^n \over n} \qty{z-a}^n \\ {\frac{\partial }{\partial z}\,} \sum_{j=0}^\infty a_j z^j &= \sum_{j=0}^\infty a_{j+1}z^j \end{align*}

• Consider $$1/f(z)$$ and $$f(1/z)$$.

# 2 Definitions

A function $$f:\Omega \to {\mathbb{C}}$$ is analytic at $$z_0\in \Omega$$ iff there exists a power series $$g(z) = \sum a_n (z-z_0)^n$$ with radius of convergence $$R>0$$ and a neighborhood $$U\ni z_0$$ such that $$f(z) = g(z)$$ on $$U$$.

\begin{align*} u_x = v_y \quad\text{and}\quad u_y = -v_x \\ \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \quad \text { and } \quad \frac{\partial v}{\partial r}=-\frac{1}{r} \frac{\partial u}{\partial \theta} \\ .\end{align*}

A function that is holomorphic on $${\mathbb{C}}$$ is said to be entire.

A singularity $$z_0$$ is essential iff it is neither removable nor a pole.

Equivalently, a Laurent series expansion about $$z_0$$ has a principal part with infinitely many terms.

A function $$f:{\mathbb{C}}\to {\mathbb{C}}$$ is holomorphic at $$z_0$$ if the following limit converges: \begin{align*} \lim_{h\to 0} {1\over h} \qty{f(z_0 + h) - f(z_0)} \mathrel{\vcenter{:}}= f'(z_0) .\end{align*}

A real function of two variables $$u(x, y)$$ is harmonic iff its Laplacian vanishes: \begin{align*} \Delta u \mathrel{\vcenter{:}}=\qty{{\frac{\partial ^2}{\partial x^2}\,} + {\frac{\partial ^2}{\partial y^2}\,}}u = 0 .\end{align*}

A function $$f:\Omega\to{\mathbb{C}}$$ is meromorphic iff there exists a sequence $$\left\{{z_n}\right\}$$ such that

• $$\left\{{z_n}\right\}$$ has no limit points in $$\Omega$$.
• $$f$$ is holomorphic in $$\Omega\setminus\left\{{z_n}\right\}$$.
• $$f$$ has poles at the points $$\left\{{z_n}\right\}$$.

If $$f$$ is either holomorphic or has a pole at $$z=\infty$$ is said to be meromorphic on $${\mathbb{CP}}^1$$.

A pole $$z_0$$ of a meromorphic function $$f(z)$$ is a zero of $$g(z) \mathrel{\vcenter{:}}={1\over f(z)}$$. If there exists an $$n$$ such that \begin{align*} \lim_{z\to z_0}\qty{z-z_0}^nf(z) \end{align*} is holomorphic and nonzero in a neighborhood of $$z_0$$, then the minimal such $$n$$ is the order of the pole. A pole of order 1 is said to be a simple pole.

The pole $$z_0$$ is isolated iff there exists a neighborhood of $$z_0$$ containing no other poles of $$f$$.

In a Laurent series $$f(z) \mathrel{\vcenter{:}}=\sum_{n\in {\mathbb{Z}}} c_n (z-z_0)^n$$, the principal part of $$f$$ at $$z_0$$ consists of terms with negative degree: \begin{align*} P_f(z) \mathrel{\vcenter{:}}=\sum_{n=1}^\infty c_{-n}(z-z_0)^{-n} .\end{align*}

The residue of $$f$$ at $$z_0$$ is the coefficient $$c_{-1}$$.

If $$z_0$$ is a singularity of $$f$$ and there exists a $$g$$ such that $$f(z) = g(z)$$ for all $$z$$ in some deleted neighborhood $$U\setminus\left\{{z_0}\right\}$$, then $$z_0$$ is a removable singularity of $$f$$.

A map of the following form is a linear fractional transformation: \begin{align*} T(z) = {az + b \over cz + d} ,\end{align*} where the denominator is assumed to not be a multiple of the numerator.

These have inverses given by \begin{align*} T^{-1}(w) = {dw-b \over -cw + a} .\end{align*}

A bijective holomorphic map is a conformal (or angle-preserving) map, a.k.a. a biholomorphism.

Note that some authors just require the weaker condition that $$f'(z) \neq 0$$ for any point.

# 3 Theorems

## 3.1 Basics

• $$f(z) = {1\over z}$$ is holomorphic on $${\mathbb{C}}\setminus\left\{{0}\right\}$$.
• $$f(z) = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu$$ is not holomorphic, since $$\mkern 1.5mu\overline{\mkern-1.5muh\mkern-1.5mu}\mkern 1.5mu \over h$$ does not converge (but is real differentiable).

If $$\Omega \subseteq {\mathbb{C}}$$ is bounded with $${{\partial}}\Omega$$ piecewise smooth and $$f, g\in C^1(\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu)$$, then \begin{align*}\int_{{{\partial}}\Omega} f\, dx + g\, dy = \iint_{\Omega} \qty{ {\frac{\partial g}{\partial x}\,} - {\frac{\partial f}{\partial y}\,} } \, dA.\end{align*}

Define the forward difference operator $$\Delta f_k = f_{k+1} - f_k$$, then \begin{align*} \sum_{k=m}^n f_k \Delta g_k + \sum_{k=m}^{n-1} g_{k+1}\Delta f_k = f_n g_{n+1} - f_m g_m \end{align*}

Note: compare to $$\int_a^b f \, dg + \int_a^b g\, df = f(b) g(b) - f(a) g(a)$$.

## 3.2 Holomorphic and Entire Functions

### 3.2.1 Key Theorems

If $$f$$ is holomorphic on $$\Omega$$, then \begin{align*} \int_{{{\partial}}\Omega} f(z) \, dz = 0 .\end{align*}

Slogan: closed path integrals of holomorphic functions vanish.

If $$f$$ is continuous on a domain $$\Omega$$ and $$\int_T f = 0$$ for every triangle $$T\subset \Omega$$, then $$f$$ is holomorphic.

Slogan: if every integral along a triangle vanishes, implies holomorphic.

If $$f$$ is holomorphic and nonconstant on an open connected region $$\Omega$$, then $${\left\lvert {f} \right\rvert}$$ can not attain a maximum on $$\Omega$$. If $$\Omega$$ is bounded and $$f$$ is continuous on $$\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu$$, then $$\max_{\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu} {\left\lvert {f} \right\rvert}$$ occurs on $${{\partial}}\Omega$$. Conversely, if $$f$$ attains a local supremum at $$z_0 \in \Omega$$, then $$f$$ is constant on $$\Omega$$.

If $$f$$ is entire and bounded, $$f$$ is constant.

Suppose $$f$$ is holomorphic on $$\Omega$$, then \begin{align*} f(z) = {1 \over 2\pi i} \oint_{{{\partial}}\Omega} {f(\xi) \over \xi-z}\,d\xi \end{align*} and \begin{align*} {\frac{\partial ^nf }{\partial z^n}\,}(z) = {n! \over 2\pi i} \int_{{{\partial}}\Omega} {f(\xi) \over (\xi - z)^{n+1}} \,d\xi .\end{align*}

For $$z_0 \in D_R(z_0) \subset \Omega$$, we have \begin{align*} {\left\lvert { f^{(n)} (z_0) } \right\rvert} \leq \frac{n !}{2 \pi} \int_{0}^{2 \pi} \frac{ {\left\lVert {f} \right\rVert}_{\infty} } {R^{n+1}} R \,d\theta = \frac{n !{\left\lVert {f} \right\rVert}_{\infty}}{R^n} ,\end{align*} where $${\left\lVert {f} \right\rVert}_{\infty}\mathrel{\vcenter{:}}=\sup_{z\in C_R} {\left\lvert {f(z)} \right\rvert}$$.

The $$n$$th Taylor coefficient of an analytic function is at most $$\sup_{{\left\lvert {z} \right\rvert} = R} {\left\lvert {f} \right\rvert}/R^n$$.

For $$f$$ meromorphic in $$\gamma^\circ$$, if $$f$$ has no poles and is nonvanishing on $$\gamma$$ then \begin{align*} \Delta_\gamma \arg f(z) = \int_\gamma {f'(z) \over f(z)} \,dz = 2\pi (Z_f - P_f) ,\end{align*} where $$Z_f$$ and $$P_f$$ are the number of zeros and poles respectively enclosed by $$\gamma$$, counted with multiplicity.

If $$f, g$$ are analytic on a domain $$\Omega$$ with finitely many zeros in $$\Omega$$ and $$\gamma \subset \Omega$$ is a closed curve surrounding each point exactly once, where $${\left\lvert {g} \right\rvert} < {\left\lvert {f} \right\rvert}$$ on $$\gamma$$, then $$f$$ and $$f+g$$ have the same number of zeros.

Alternatively:

Suppose $$f = g + h$$ with $$g \neq 0, \infty$$ on $$\gamma$$ with $${\left\lvert {g} \right\rvert} > {\left\lvert {h} \right\rvert}$$ on $$\gamma$$. Then \begin{align*}\Delta_\gamma \arg(f) = \Delta_\gamma \arg(h)\quad\text{ and } Z_f - P_f = Z_g - P_g.\end{align*}

If $$f$$ is holomorphic on an open set $$\Omega$$ containing a curve $$\gamma$$ and its interior $$\gamma^\circ$$, except for finitely many poles $$\left\{{z_k}\right\}_{k=1}^N \subset \gamma^\circ$$. Then \begin{align*} \int_\gamma f(z) \,dz = 2\pi i \sum_{k=1}^N \mathop{\mathrm{Res}}_{z_k} f .\end{align*}

The fractional linear transformation given by $$F(z) = {i - z \over i + z}$$ maps $${\mathbb{D}}\to {\mathbb{H}}$$ with inverse $$G(w) = i {1-w \over 1 + w}$$.

If $$f: {\mathbb{D}}\to {\mathbb{D}}$$ is holomorphic with $$f(0) = 0$$, then

1. $${\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}$$ for all $$z\in {\mathbb{D}}$$
2. $${\left\lvert {f'(0)} \right\rvert} \leq 1$$.

Moreover, if $${\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {z_0} \right\rvert}$$ for any $$z_0\in {\mathbb{D}}$$ or $${\left\lvert {f'(0)} \right\rvert} = 1$$, then $$f$$ is a rotation

\begin{align*} f(z_0) = {1\over \pi r^2} \iint_{D_r(z_0)} f(z)\, dA .\end{align*}

If $$f$$ is continuous and holomorphic on $${\mathbb{H}}^+$$ and real-valued on $${\mathbb{R}}$$, then the extension defined by $$F(z) = \mkern 1.5mu\overline{\mkern-1.5muf(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu$$ for $$z\in {\mathbb{H}}^-$$ is a well-defined holomorphic function on $${\mathbb{C}}$$.

There are 8 major types of conformal maps:

Type/Domains Formula
Translation/Dilation/Rotation $$z\mapsto e^{i\theta}(cz + h)$$
Sectors to sectors $$z\mapsto z^n$$
$${\mathbb{D}}_{1\over 2} \to {\mathbb{H}}_{1\over 2}$$, the first quadrant $$z\mapsto {1+z \over 1-z}$$
$${\mathbb{H}}\to S$$ $$z\mapsto \log(z)$$
$${\mathbb{D}}_{1\over 2} \to S_{1\over 2}$$ $$z\mapsto \log(z)$$
$$S_{1\over 2} \to {\mathbb{D}}_{1\over 2}$$ $$z\mapsto e^{iz}$$
$${\mathbb{D}}_{1\over 2} \to {\mathbb{H}}$$ $$z\mapsto {1\over 2}\qty{z + {1\over z}}$$
$$S_{1\over 2} \to {\mathbb{H}}$$ $$z\mapsto \sin(z)$$

Conformal maps $${\mathbb{D}}\to{\mathbb{D}}$$ have the form \begin{align*} g(z) = \lambda {1-a \over 1 - \mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu z}, \quad {\left\lvert {a} \right\rvert} < 1, \quad {\left\lvert {\lambda} \right\rvert} = 1 .\end{align*}

### 3.2.2 Others

If $$\Omega$$ is simply connected, nonempty, and not $${\mathbb{C}}$$, then for every $$z_{0}\in \Omega$$ there exists a unique conformal map $$F:\Omega \to {\mathbb{D}}$$ such that $$F(z_{0}) = 0$$ and $$F'(z_{0}) > 0$$.

Thus any two such sets $$\Omega_{1}, \Omega_{2}$$ are conformally equivalent.

If $$f$$ is holomorphic on $$\Omega$$ except possibly at $$z_0$$ and $$f$$ is bounded on $$\Omega\setminus\left\{{z_0}\right\}$$, then $$z_0$$ is a removable singularity.

If $$f(z) = u(x, y) + iv(x, y)$$ is holomorphic, then $$u, v$$ are harmonic.

$$f$$ is holomorphic at $$z_0$$ iff there exists an $$a\in {\mathbb{C}}$$ such that \begin{align*} f(z_0 + h) - f(z_0) - ah = h \psi(h), \quad \psi(h) \overset{h\to 0}\to 0 .\end{align*} In this case, $$a = f'(z_0)$$.

If $$f = u+iv$$ with $$u, v\in C^1({\mathbb{R}})$$ satisfying the Cauchy-Riemann equations on $$\Omega$$, then $$f$$ is holomorphic on $$\Omega$$ and $$f'(z) = {\frac{\partial f}{\partial z}\,} = {1 \over 2} \qty{{\frac{\partial }{\partial x}\,} + {1\over i} {\frac{\partial }{\partial y}\,}}f$$.

\begin{align*} \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \quad \text { and } \quad \frac{1}{r} \frac{\partial u}{\partial \theta}=-\frac{\partial v}{\partial r} .\end{align*}

(Click to expand)
• Take derivative along two paths, along a ray with constant angle $$\theta_0$$ and along a circular arc of constant radius $$r_0$$.
• Then equate real and imaginary parts.

Any holomorphic non-constant map is an open map.

## 3.3 Series and Analytic Functions

Any power series is smooth and its derivatives can be obtained using term-by-term differentiation.

A series of functions $$\sum_{n=1}^\infty f_n(x)$$ converges uniformly iff \begin{align*} \lim_{n\to \infty} {\left\lVert { \sum_{k\geq n} f_k } \right\rVert}_\infty = 0 .\end{align*}

If $$\left\{{f_n}\right\}$$ with $$f_n: \Omega \to {\mathbb{C}}$$ and there exists a sequence $$\left\{{M_n}\right\}$$ with $${\left\lVert {f_n} \right\rVert}_\infty \leq M_n$$ and $$\sum_{n\in {\mathbb{N}}} M_n < \infty$$, then $$f(x) \mathrel{\vcenter{:}}=\sum_{n\in {\mathbb{N}}} f_n(x)$$ converges absolutely and uniformly on $$\Omega$$.

Moreover, if the $$f_n$$ are continuous, by the uniform limit theorem, $$f$$ is again continuous.

$$f(z) = e^z$$ is uniformly convergent in any disc in $${\mathbb{C}}$$.

(Click to expand)

Apply the estimate \begin{align*} {\left\lvert {e^z} \right\rvert} \leq \sum {{\left\lvert {z} \right\rvert}^n \over n!} = e^{{\left\lvert {z} \right\rvert}} .\end{align*} Now by the $$M{\hbox{-}}$$test, \begin{align*} {\left\lvert {z} \right\rvert} \leq R < \infty \implies {\left\lvert {\sum {z^n \over n!}} \right\rvert} \leq e^R < \infty .\end{align*}

For a power series $$f(z) = \sum a_n z^n$$, define $$R$$ by \begin{align*} {1\over R}\mathrel{\vcenter{:}}=\limsup {\left\lvert {a_n} \right\rvert}^{1\over n} .\end{align*}

Then $$f$$ converges absolutely on $${\left\lvert {z} \right\rvert} < R$$ and diverges on $${\left\lvert {z} \right\rvert} > R$$.

## 3.4 Others

If $$f$$ is holomorphic on $$\Omega\setminus\left\{{z_0}\right\}$$ where $$z_0$$ is an essential singularity, then for every $$V\subset \Omega\setminus\left\{{z_0}\right\}$$, $$f(V)$$ is dense in $${\mathbb{C}}$$.

The image of a disc punctured at an essential singularity is dense in $${\mathbb{C}}$$.

If $$f:{\mathbb{C}}\to {\mathbb{C}}$$ is entire and nonconstant, then $$\operatorname{im}(f)$$ is either $${\mathbb{C}}$$ or $${\mathbb{C}}\setminus\left\{{z_0}\right\}$$ for some point $$z_0$$.

If $$f$$ is holomorphic on a bounded connected domain $$\Omega$$ and there exists a sequence $$\left\{{z_i}\right\}$$ with a limit point in $$\Omega$$ such that $$f(z_i) = 0$$, then $$f\equiv 0$$ on $$\Omega$$.

Two functions agreeing on a set with a limit point are equal on a domain.

The ring of holomorphic functions on a domain in $${\mathbb{C}}$$ has no zero divisors.

(Click to expand)

???

If $$z_0$$ is a zero of $$f'$$ of order $$n$$, then $$f$$ is $$(n+1)$$-to-one in a neighborhood of $$z_0$$.

(Click to expand)

?

For $$\Omega\subseteq{\mathbb{C}}$$, show that $$A({\mathbb{C}})\mathrel{\vcenter{:}}=\left\{{f: \Omega \to {\mathbb{C}}{~\mathrel{\Big|}~}f\text{ is bounded}}\right\}$$ is a Banach space.

(Click to expand)

?

Apply Morera’s Theorem and Cauchy’s Theorem

# 4 Residues

Check: do you need residues? You may be able to just compute an integral

• Directly by parameterization: \begin{align*} \int_\gamma f = \int_a^b f(z(t))\, z'(t) && \text{for } z(t) \text{ a parameterization of } \gamma ,\end{align*}

• Finding a primitive $$F$$,

• Writing $$z= z_0 + re^{i \theta }$$

(Click to expand)
• Given $$z_0\in \Omega$$, pick the largest disc $$D_R(z_0) \subset \Omega$$ and let $$C_R = {{\partial}}D_R$$.
• Then apply the integral formula.

If $$z_0$$ is a simple pole of $$f$$, then \begin{align*} \mathop{\mathrm{Res}}_{z_0}f = \lim_{z\to z_0} (z-z_0) f(z) .\end{align*}

Let $$f(z) = \frac{1}{1+z^2}$$, then $$\mathop{\mathrm{Res}}(i, f) = \frac{1}{2i}$$.

If $$f$$ has a pole $$z_0$$ of order $$n$$, then \begin{align*} \mathop{\mathrm{Res}}_{z=z_0} f = \lim_{z\to z_0} {1 \over (n-1)!} \qty{{\frac{\partial }{\partial z}\,}}^{n-1} (z-z_0)^n f(z) .\end{align*}

# 5 Conformal Maps

A bijective holomorphic map automatically has a holomorphic inverse. This can be weakened: an injective holomorphic map satisfies $$f'(z) \neq 0$$ and $$f ^{-1}$$ is well-defined on its range and holomorphic.

## 5.1 Plane to Disc

\begin{align*} \phi: {\mathbb{H}}&\to {\mathbb{D}}\\ \phi(z) &= {z - i \over z + i} \qquad f^{-1}(z) = i\qty{1 + w \over 1 - w} .\end{align*}

## 5.2 Sector to Disc

For $$S_\alpha \mathrel{\vcenter{:}}=\left\{{z\in{\mathbb{C}}{~\mathrel{\Big|}~}0 < \arg(z) < \alpha }\right\}$$ an open sector for $$\alpha$$ some angle, first map the sector to the half-plane: \begin{align*} g: S_\alpha &\to {\mathbb{H}}\\ g(z) &= z^{\pi \over \alpha} .\end{align*}

Then compose with a map $${\mathbb{H}}\to{\mathbb{D}}$$: \begin{align*} f: S_\alpha &\to {\mathbb{D}}\\ f(z) &= (\phi \circ g)(z) = {z^{\pi\over \alpha} - i \over z^{\pi\over\alpha} + i} .\end{align*}

## 5.3 Strip to Disc

• Map to horizontal strip by rotation $$z\mapsto \lambda z$$.
• Map horizontal strip to sector by $$z \mapsto e^z$$
• Map sector to $${\mathbb{H}}$$ by $$z\mapsto z^{\pi\over\alpha}$$.
• Map $${\mathbb{H}}\to{\mathbb{D}}$$.

# 6 Schwarz Reflection

$${\mathbb{H}}^+, {\mathbb{H}}^-$$ can be replaced with any region symmetric about a line segment $$L\subseteq {\mathbb{R}}$$.

# 7 Zeros and Poles

## 7.1 Counting Zeros

• Take $$P(z) = z^4 + 6z + 3$$.
• On $${\left\lvert {z} \right\rvert} < 2$$:
• Set $$f(z) = z^4$$ and $$g(z) = 6z + 3$$, then $${\left\lvert {g(z)} \right\rvert} \leq 6{\left\lvert {z} \right\rvert} + 3 = 15 < 16= {\left\lvert {f(z)} \right\rvert}$$.
• So $$P$$ has 4 zeros here.
• On $${\left\lvert {z} \right\rvert} < 1$$:
• Set $$f(z) = 6z$$ and $$g(z) = z^4 + 3$$.
• Check $${\left\lvert {g(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^4 + 3 = 4 < 6 = {\left\lvert {f(z)} \right\rvert}$$.
• So $$P$$ has 1 zero here.
• Claim: the equation $$\alpha z e^z = 1$$ where $${\left\lvert {\alpha} \right\rvert} > e$$ has exactly one solution in $${\mathbb{D}}$$.
• Set $$f(z) = \alpha z$$ and $$g(z) = e^{-z}$$.
• Estimate at $${\left\lvert {z} \right\rvert} =1$$ we have $${\left\lvert {g} \right\rvert} ={\left\lvert {e^{-z}} \right\rvert} = e^{-\Re(z)} \leq e^1 < {\left\lvert {\alpha} \right\rvert} = {\left\lvert {f(z)} \right\rvert}$$
• $$f$$ has one zero at $$z_0 = 0$$, thus so does $$f+g$$.

# 9 Appendix: Proofs of the Fundamental Theorem of Algebra

### 9.0.1 Fundamental Theorem of Algebra: Argument Principle

• Let $$P(z) = a_nz^n + \cdots + a_0$$ and $$g(z) = P'(z)/P(z)$$, note $$P$$ is holomorphic
• Since $$\lim_{{\left\lvert {z} \right\rvert} \to \infty} P(z) = \infty$$, there exist an $$R>0$$ such that $$P$$ has no roots in $$\left\{{{\left\lvert {z} \right\rvert} \geq R}\right\}$$.
• Apply the argument principle: \begin{align*} N(0) = {1\over 2\pi i} \oint_{{\left\lvert {\xi} \right\rvert} = R} g(\xi) \,d\xi .\end{align*}
• Check that $$\lim_{{\left\lvert {z\to \infty} \right\rvert}}zg(z) = n$$, so $$g$$ has a simple pole at $$\infty$$
• Then $$g$$ has a Laurent series $${n\over z} + {c_2 \over z^2} + \cdots$$
• Integrate term-by-term to get $$N(0) = n$$.

### 9.0.2 Fundamental Theorem of Algebra: Rouche’s Theorem

• Let $$P(z) = a_nz^n + \cdots + a_0$$
• Set $$f(z) = a_n z^n$$ and $$g(z) = P(z) - f(z) = a_{n-1}z^{n-1} + \cdots + a_0$$, so $$f+g = P$$.
• Choose $$R > \max\qty{ { {\left\lvert {a_{n-1}} \right\rvert} + \cdots + {\left\lvert {a_0} \right\rvert} \over {\left\lvert {a_n} \right\rvert} }, 1}$$, then

\begin{align*} |g(z)| &\coloneqq|a_{n-1}z^{n-1} + \cdots + a_1 z + a_0 | \\ &\leq |a_{n-1}z^{n-1}| + \cdots + |a_1 z| + |a_0 | \quad\text{by the triangle inequality} \\ &= |a_{n-1}|\cdot |z^{n-1}| + \cdots + |a_1|\cdot| z| + |a_0 | \\ &= |a_{n-1}|\cdot R^{n-1} + \cdots + |a_1| R + |a_0 | \\ &\leq |a_{n-1}|\cdot R^{n-1}+|a_{n-2}|\cdot R^{n-1} + \cdots + |a_1| \cdot R^{n-1} + |a_0 |\cdot R^{n-1} \quad\text{since } R>1 \implies R^{a+b} \geq R^a \\ &= R^{n-1} \left( |a_{n-1}| + |a_{n-2}| + \cdots + |a_1| + |a_0| \right) \\ &\leq R^{n-1} \left( |a_n|\cdot R \right) \quad\text{by choice of } R \\ &= R^{n} |a_n| \\ &= |a_n z^n| \\ &\coloneqq|f(z)| \end{align*}

• Then $$a_n z^n$$ has $$n$$ zeros in $${\left\lvert {z} \right\rvert} < R$$, so $$f+g$$ also has $$n$$ zeros.

### 9.0.3 Fundamental Theorem of Algebra: Liouville’s Theorem

• Suppose $$p$$ is nonconstant and has no roots, then $${1\over p}$$ is entire. We will show it is also bounded and thus constant, a contradiction.
• Write $$p(z) = z^n \left(a_n + \frac{a_{n-1}}{z}+\dots+\frac{a_{0}}{z^{n}}\right)$$
• Outside a disc:
• Note that $$p(z) \overset{z\to \infty }\to \infty$$. so there exists an $$R$$ large enough such that $${\left\lvert {p(z)} \right\rvert} \geq {1\over A}$$ for any fixed chosen constant $$A$$.
• Then $${\left\lvert { 1/p(z)} \right\rvert} \leq A$$ outside of $${\left\lvert {z} \right\rvert} >R$$, i.e. $$1/p(z)$$ is bounded there.
• Inside a disc:
• $$p$$ is continuous with no roots and thus must be bounded below on $${\left\lvert {z} \right\rvert} < R$$.
• $$p$$ is entire and thus continuous, and since $$\mkern 1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu_r(0)$$ is a compact set, $$p$$ achieves a min $$A$$ there
• Set $$C \mathrel{\vcenter{:}}=\min(A, B)$$, then $${\left\lvert {p(z)} \right\rvert} \geq C$$ on all of $${\mathbb{C}}$$ and thus $${\left\lvert {1/p(z)} \right\rvert} \leq C$$ everywhere.
• So $$1/p(z)$$ is bounded an entire and thus constant by Liouville’s theorem – but this forces $$p$$ to be constant. $$\contradiction$$

### 9.0.4 Fundamental Theorem of Algebra: Open Mapping Theorem

• $$p$$ induces a continuous map $${\mathbb{CP}}^1 \to {\mathbb{CP}}^1$$
• The continuous image of compact space is compact;
• Since the codomain is Hausdorff space, the image is closed.
• $$p$$ is holomorphic and non-constant, so by the Open Mapping Theorem, the image is open.
• Thus the image is clopen in $${\mathbb{CP}}^1$$.
• The image is nonempty, since $$p(1) = \sum a_i \in {\mathbb{C}}$$
• $${\mathbb{CP}}^1$$ is connected
• But the only nonempty clopen subset of a connected space is the entire space.
• So $$p$$ is surjective, and $$p^{-1}(0)$$ is nonempty.
• So $$p$$ has a root.

# 10 Appendix

## 10.1 Misc Basic Algebra

• Circle: $$x^2 + y^2 = r^2$$
• Ellipse: $$\qty{\frac x a}^2 + \qty{\frac y b}^2 = 1$$
• Hyperbola: $$\qty{\frac x a}^2 - \qty{\frac y b}^2 = 1$$
• Rectangular Hyperbola: $$xy = \frac{c^2}{2}$$.
• Parabola: $$-4ax + y^2 = 0$$.

Mnemonic: Write $$f(x, y) = Ax^2 + Bxy + Cy^2 + \cdots$$, then consider the discriminant $$\Delta = B^2 - 4AC$$:

• $$\Delta < 0 \iff$$ ellipse
• $$\Delta < 0$$ and $$A=C, B=0 \iff$$ circle
• $$\Delta = 0 \iff$$ parabola
• $$\Delta > 0 \iff$$ hyperbola

\begin{align*} x^2 - bx = (x - s)^2 - s^2 \quad\text{where} s = \frac{b}{2} \\ x^2 + bx = (x + s)^2 - s^2 \quad\text{where} s = \frac{b}{2} .\end{align*}

The sum of the interior angles of an $$n{\hbox{-}}$$gon is $$(n-2)\pi$$, where each angle is $$\frac{n-2}{n}\pi$$.

Basics

• Show that $${1\over z} \sum_{k=1}^\infty {z^k \over k}$$ converges on $$S^1 \setminus\left\{{1}\right\}$$ using summation by parts.
• Show that any power series is continuous on its domain of convergence.
• Show that a uniform limit of continuous functions is continuous.

??

• Show that if $$f$$ is holomorphic on $${\mathbb{D}}$$ then $$f$$ has a power series expansion that converges uniformly on every compact $$K\subset {\mathbb{D}}$$.

• Show that any holomorphic function $$f$$ can be uniformly approximated by polynomials.

• Show that if $$f$$ is holomorphic on a connected region $$\Omega$$ and $$f'\equiv 0$$ on $$\Omega$$, then $$f$$ is constant on $$\Omega$$.

• Show that if $${\left\lvert {f} \right\rvert} = 0$$ on $${{\partial}}\Omega$$ then either $$f$$ is constant or $$f$$ has a zero in $$\Omega$$.

• Show that if $$\left\{{f_n}\right\}$$ is a sequence of holomorphic functions converging uniformly to a function $$f$$ on every compact subset of $$\Omega$$, then $$f$$ is holomorphic on $$\Omega$$ and $$\left\{{f_n'}\right\}$$ converges uniformly to $$f'$$ on every such compact subset.

• Show that if each $$f_n$$ is holomorphic on $$\Omega$$ and $$F \mathrel{\vcenter{:}}=\sum f_n$$ converges uniformly on every compact subset of $$\Omega$$, then $$F$$ is holomorphic.

• Show that if $$f$$ is once complex differentiable at each point of $$\Omega$$, then $$f$$ is holomorphic.

# 11 Draft of Problem Book

• Prove the triangle inequality
• Prove the reverse triangle inequality
• Show that $$\sum z^{k-1}/k$$ converges for all $$z\in S^1$$ except $$z=1$$.
• What is an example of a noncontinuous limit of continuous functions?
• Show that the uniform limit of continuous functions is continuous.
• Show that $$f$$ is holomorphic if and only if $$\mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5muf = 0$$.
• Show $$n^{1\over n} \overset{n\to \infty } \to 1$$.
• Show that if $$f$$ is holomorphic with $$f'=0$$ on $$\Omega$$ then $$f$$ is constant.
• Show that holomorphic implies analytic.
• Use Cauchy’s inequality to prove Liouville’s theorem

What is a pair of conformal equivalences between $${\mathbb{H}}$$ and $${\mathbb{D}}$$?

(Click to expand)

\begin{align*} F: HH &\to {\mathbb{D}}\\ z & \mapsto {i-z \over i+z} \\ \\ G: {\mathbb{D}}&\to {\mathbb{H}}\\ w &\mapsto i{1-w \over 1 + w} .\end{align*}

Mnemonic: any point in $${\mathbb{H}}$$ is closer to $$i$$ than $$-i$$, so $${\left\lvert {F(z)} \right\rvert} < 1$$.

• Maps $${\mathbb{R}}\to S^1\setminus\left\{{-1}\right\}$$.

What is conformal equivalence $${\mathbb{H}}\rightleftharpoons S \mathrel{\vcenter{:}}=\left\{{w\in {\mathbb{C}}{~\mathrel{\Big|}~}0 < \arg(w) < \alpha \pi}\right\}$$?

(Click to expand)

\begin{align*} f(z) = z^ \alpha .\end{align*}