# Math416 Lecture 16

## Answer checking for exam

### Q1

Cauchy riemann equations:

$$
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\quad\text{and}\quad\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
$$

Liouville's Theorem:

Any non-constant entire function is unbounded.

So $\cos(z)$ is unbounded in $\mathbb{C}$.

$$
\text{Log}(-e^2) = \ln|-e^2| + i\arg(-e^2) = -2 + \pi i
$$

At any point $z_0\in \mathbb{C}\setminus\{0\}$, there is an open set $z_0\in U\subset \mathbb{C}$ and a branch of logarithm defined on $U$.

### Q2

Power series expansion

### Q3

limit superior

### Q4

Bound integral

### Q5

$f_n$ converges pointwise to $f$ on $U$ if $\forall z\in U$, $\forall \epsilon > 0$, $\exists N$ s.t. $\forall n\geq N$, $|f_n(z)-f(z)| < \epsilon$.

$f_n$ converges uniformly to $f$ on $U$ if $\forall \epsilon > 0$, $\exists N$ s.t. $\forall n\geq N$, $\forall z\in U$, $|f_n(z)-f(z)| < \epsilon$.

Show for $|z|<1$, $f_n(z)=z^n$ converges pointwise to $0$ but not uniformly to $0$.

(a) pointwise convergence:

$|z^n| = |z|^n < \epsilon$ if $n > \frac{\ln\epsilon}{\ln|z|}$.

(b) uniform convergence:

No matter how small $\epsilon$ is, there is always a $z$ s.t. $|z^n| > \epsilon$ for all $n$.

## Continue from last lecture

### Schwarz's Lemma

Let $f$ be an holomorphic function that maps the unit disk $D(0,1)$ to itself and $f(0)=0$. Then $|f(z)|\leq |z|$ for all $z\in D(0,1)$

#### Schwarz-Pick's Lemma

(see exercise 7.17.2)

Let $f$ be an holomorphic function that maps the unit disk $D(0,1)$ to itself. Then $\forall z,w\in D(0,1)$,

$$
\left|\frac{f(z)-f(w)}{1-\overline{f(w)}f(z)}\right|\leq \left|\frac{z-w}{1-\overline{w}z}\right|
$$

> Recall the Möbius map
>
> $$\phi_\alpha(z) = \frac{z-\alpha}{1-\overline{\alpha}z}$$
>
> is a homeomorphism of the unit disk.
> 
> So we can use the Möbius to restate the Schwarz-Pick's Lemma as:
>
> $$|\phi_{f(w)}(f(z))|\leq |\phi_w(z)|$$

Suppose we defined $g=\phi_{f(w)}\circ f\circ \phi_{-w}$, then $g$ is a holomorphic function that maps the unit disk to itself and $g(0)=0$.

By Schwarz's Lemma, let $z\in D(0,1)$, $|g(z)|\leq |z|$.

$$
|\phi_{f(w)}(f(\phi_{-w}(z)))|\leq |z|
$$

Let $\zeta=\phi_{-w}(z)$, then $\zeta=\frac{z+w}{1+\overline{w}z}\in D(0,1)$, so $|\zeta|=\phi_w(z)$.

#### Extension of Schwarz-Pick's Lemma in hyperbolic metric

Suppose we defined the distance on $\mathbb{C}$ as $d(z,w)=|\frac{z-w}{1-\overline{w}z}|$.

We claim that this is a metric on $\mathbb{C}$. $\forall z,w,v\in \mathbb{C}$:

(a) $d(z,w)=0$ if and only if $z=w$ and $d(z,w)> 0$ otherwise.

(b) $d(z,w)=d(w,z)$.

(c) $d(z,w)\leq d(z,v)+d(v,w)$.

We call this metric the Pseudo hyperbolic metric.

> Hyperbolic metric:
>
> $$ \text{Hypdist}(z,w)=\tanh^{-1}(d(z,w))$$
> 
> Where $d(z,w)=|\frac{z-w}{1-\overline{w}z}|$

So we can restate the Schwarz-Pick's Lemma as:

$$
d(f(z),f(w))\leq d(z,w)
$$

And in hyperbolic metric, it becomes:

$$
\text{Hypdist}(f(z),f(w))\leq \text{Hypdist}(z,w)
$$

Suppose the equality holds for Schwarz-Pick's Lemma, then $|g(z)|=\tau z$ where $|\tau|=1$.

Computation ignored here.

Then $f$ is a Möbius map that is automorphism of the unit disk.

### Existence of harmonic conjugate

Suppose $f=u+iv$ is holomorphic on a domain $U\subset \mathbb{C}$. Then $u=\text{Re}(f)$ is harmonic on $U$. That is $\Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$.

#### Theorem 7.18

Let $u$ be a real harmonic function on a convex domain $G\subset \mathbb{C}$. Then there exists $g\in O(G)$ such that $\text{Re}(g)=u$. Moreover, $g$ is unique up to an additive imaginary constant.

Proof:

Existence next time.

Uniqueness:

Suppose $g,h\in O(G)$ s.t. $\text{Re}(g)=\text{Re}(h)=u$.

$\text{Re}g=u=\text{Re}h$ on $G$.

If we can show that $(g-h)'=0$ on $G$, then we win.

Let $g=u+iv$, $h=u+iw$.

By the Cauchy-Riemann equations,

$$
\begin{aligned}
\frac{\partial}{\partial x}(g-h)&=\frac{\partial}{\partial x}i(v-w)\\
&=i\left(\frac{\partial u}{\partial y}-\frac{\partial u}{\partial y}\right)\\
&=0
\end{aligned}
$$

Suppose $G=\mathbb{C}\setminus\{0\}$, then $u=\ln|z|=\frac{1}{2}\ln(x^2+y^2)$, which is harmonic.

Continue next time.