Problem $\# 1$

Recall that $\cos z=\frac{e^{i z}+e^{-i z}}{2},$ for $z=x+i y \in \mathcal{C}$

(a) Show that $\cos z$ reduces to $\cos x$ when $z=x \in \mathcal{R}$.

Solution:

$e^{i x}=\cos x+i \sin x,$ so
$$\frac{e^{i x}+e^{-i x}}{2}=\frac{\cos x+i \sin x+\cos (-x)+i \sin (-x)}{2}=\frac{2 \cos x}{2}=\cos x$$

(b) Find $u(x, y)$ and $v(x, y)$ real so that $\cos (z)=u(x, y)+i v(x, y)$.\\

Solution:
\begin{aligned} \cos (x+i y) &=\frac{e^{i x-y}+e^{-i x+y}}{2}=\frac{e^{-y}(\cos x+i \sin x)+e^{y}(\cos (-x)+i \sin (-x))}{2} \\ &=\frac{e^{y}+e^{-y}}{2} \cos x-\frac{e^{y}-e^{-y}}{i 2} \sin x \end{aligned}
so
$$u=\cosh (y) \cos (x), \quad v=-\sinh (y) \sin (x)$$

(c) Prove $f(z)=\cos z$ satisfies the Cauchy-Riemann equations $u_{x}=v_{y}$
$$u_{y}=-v_{x}$$
Solution:
$$\begin{array}{l} u_{x}=-\cosh (y) \sin (x), \quad v_{y}=-\cosh (y) \sin (x) \\ u_{y}=\sinh (y) \cos (x), \quad v_{x}=-\sinh (y) \cos (x) \end{array}$$

Problem $\# 2$ Solution

$$2^{1 / \pi}=e^{\frac{1}{\pi} \log (i)}=e^{\frac{1}{\pi} i(\pi / 2+2 k \pi)}=e^{i\left(\frac{1}{2}+2 k\right)}$$

Claim: There are all distinct roots.
I.e. , if not, then $\frac{1}{2}+2 k_{1}=\frac{1}{2}+2 k_{2}+2 m \pi$
for some integers $k_{1}, k_{2}, m$. But then
$$\frac{k_{1}-k_{2}}{m}=\pi \Rightarrow \pi \text { rational. }$$

(a)

$$f^{\prime}(z)=\frac{1}{2 \pi i} \int_{C_{R}} \frac{f(w)}{(w-z)^{2}} d w$$

$$\left|f^{\prime}(z)\right| \leqslant\left|\frac{1}{2 \pi}\right|\left|C_{R}\right| \begin{array}{c} \operatorname{Max} \\ C_{B} \end{array}\left|\frac{f(w)}{(W z^{\prime})^{2}}\right|$$

$$\begin{array}{l} M=\operatorname{Max}|f| \\ \qquad \leqslant\left|\frac{1}{2 \pi}\right| 2 \pi R \frac{M}{R^{2}}=\frac{M}{R} \end{array}$$

(b)

Assume $|f(z)| \leqslant M$. Then
$\left|f^{\prime}(z)\right| \leqslant \frac{M}{R}$ holds $\forall R>0$, every $z \in$
$\left|f^{\prime}(t)\right|=0 \Rightarrow f^{\prime}(z)=0 \Rightarrow f(z)=constant$

Proof .

(a)

C-R equation

$\begin{array}{lll}u_{x}=v_{y} & u_{x x}=v_{x y} & v_{x x}=-u_{y x} \\ v_{x}=-u_{y} & u_{y y}=-v_{x y} & V_{y y}=u_{x y}\end{array}$.

$U_{x x+} V_{y y=0} \quad V_{x x}+V_{y y}=0$.

(b)

\begin{aligned} f(0) &=\frac{1}{2 \pi_{i}} \int_{e_{1}} \frac{f(w)}{w} d w \\ &=\frac{1}{2 \pi i} \int_{0}^{2 \pi} \frac{f\left(e^{i t}\right)}{e^{i t}} i e^{i t} d t \\ &=\frac{1}{2 \pi} \int_{0}^{2 \pi} f\left(e^{i t}\right) d t \end{aligned}

$$\begin{array}{l} u(0)+i v(0)=\frac{1}{2 \pi} \int_{0}^{2 \pi} u\left(e^{i t}\right) d t+\frac{i}{2 \pi} \int_{0}^{2 \pi} V\left(e^{i t}\right) d t \\ \Rightarrow u(0)=\frac{1}{2 \pi} \int_{0}^{2 \pi} u\left(e^{i t}\right) d t \end{array}$$

Proof .

(a) $P(z)=a_{n} z^{n}+\cdots+a_{0}$

$$|P(z)| \geq\left|a_{n} z^{n}\right|-\left|a_{n-1} z^{n-1}+\cdots+a_{0}\right|$$

Letting $z+e^{i \theta}$ giver $(w \log r \geq 1)$.

\begin{aligned}

|P(z)| & \geq\left|a_{A}\right| r^{n}-\left(\left|a_{n-1}\right| r^{n-1}+\cdots+\left|a_{0}\right|\right.\\

& \geq 2 \\

\text { if } r &>\frac{\left|a_{n-1}\right| r^{n-1}+\cdots \cdot \psi\left|a_{0}\right|}{\left|a_{n}\right| r^{n-1}}+2 \\

\text { so } r &\left.>\frac{\left|a_{n-1}\right|+\cdots+\left|a_{0}\right|}{\left|a_{n}\right|}+2\right\rangle 1\frac{\left|a_{n-1}\right| r_{+}^{n-1}+a_{n} \mid}{\left|a_{n}\right| r^{n-1}}+2

\end{aligned}

(b)

Assume $P(z) \neq 0$. Then

$$f(z)=\frac{1}{|P(t)|} \text { is entire }$$

But by (a)

$\exists R \leqslant|z| \geq R$ implies

$$|P(z)| \leqslant 2$$

So

$$\frac{1}{|P(z)|} \leqslant \frac{1}{2} .$$

$|P(t)|^{-1} \operatorname{cons} t \Rightarrow|P(z)|^{-1}$ takes a max

value on $\overline{B_{R}(0)}$ compact. Thus $\frac{1}{|P(\tau)|} \leqslant \operatorname{Max}\left\{\frac{1}{2}, M\right\}$

$\Rightarrow \frac{1}{\mid p(z) \mid}$ bounded entire $\Rightarrow$ const.

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