Cana Assignment 4

Name: Shubh Sharma Roll Number: BMC202170

1. PG 130, Problem 5

If $f$ has an essential singularity, it would come arbitrarily close to every point, then $f$ is would not be bounded, hence $f$ does not have an essential singularity.

Let $z_0$ be a pole of $f$, then by the open mapping theorem, an open set around $z_0$ would be mapped to an open neighbourhood of $\infty$, which would mean, either the $Re$ or the $Im$ parts are unbounded, Hence $f$ does not have a pole either.

Hence an isolated singularity of $f(z)$ is removable.

2. PG 130, Problem 6

If $e^{f(z)}$ has a pole at $z_0$ then the real part of $f(z)$ goes to $\infty$ as $z$ goes to $z_0$. Hence $f(z)$ would also have a pole at $z_0$.

Having a pole at $z_0$ implies, given any $M>0,\exists ϵ>0$ Such that $z\in B_{ϵ}(z_0)⟹|f(z_0)|>M$. Pick $z\in B_{ϵ}(z_0)\ne z_0$, its image is finite and the sequence $f(z)+2n\pi i$ converges to $\infty$., take the inverse images of each of the points of the sequence. which form a sequence $z_n\to z_0$. $e^{f(z_n)}$ is a constant sequence which is finite, hence $z_0$ is not a pole of $e^{f(z)}$ which is a contradiction.

3. PG 133, Problem 3

Let $f(z)=\cos z$ and let $z_0=0$, $f(z)-1$ has a root at $z_0$ The derivative of $f^{\prime }(0)=\sin 0=0$, second derivative is $-\cos 0=-1$. Hence the root has degree $2$. Hence $\cos z-1=z^{2}g(z)={\zeta }^2(z)$ , hence $\zeta (z)=z\sqrt{g(z)}$ and we get

$$\begin{array}{rl}f(z)-1& =\cos z-1\\ & =-2{\sin }^2\left(\frac{z}{2}\right)\\ & ={\zeta }^2(z)\\ \therefore \zeta (z)& =\sqrt{2}i\sin \left(\frac{z}{2}\right)\end{array}$$

4. PG 133, Problem 4

$f(z^n)-f(0)$ has all $i$ order derivatives for $i\in \{1\dots n-1\}$ and $0$ is a root hence $f(z^n)-f(0)=z^{n}h(z)$ where $h(0)\ne 0$ Hence we can choose a neighborhood of $0$ such that $|h(z)-h(0)|\le |h(0)|$ for all $z$ in the neighborhood, so we can define a single valued branch of $\sqrt[n]{h(z)}$ in the neighbourhood of $0$

$$f(z^n)-f(0)={\left(z\sqrt[n]{h(z)}\right)}^n$$

and have $g(z)=z\sqrt[n]{h(z)}$

5. PG 136, Problem 1

The equation $36$ in ahlfors states that, given a function $f$ analytic on $|z|<R$, such taht $|f(z)|<M,f(z_0)=w_0$, then we have

$$\left|\frac{M(f(z)-w_0)}{M^2-{\bar{w}}_{0}f(z)}\right|\le \left|\frac{R(z-z_0)}{R^2-{\bar{z}}_{0}z}\right|$$

putting $R=M=1$ we get

$$\left|\frac{f(z)-w_0}{1-{\bar{w}}_{0}f(z)}\right|\le \left|\frac{z-z_0}{1-{\bar{z}}_{0}z}\right|$$

as $z\to z_0$ we have $f(z)\to w_0$

$$\left|\frac{f^{\prime }(z_0)}{1-|f^2(z_0)}\right|\le \left|\frac{1}{1-|z_0{|}^2}\right|$$

but $z_0$ can be any point

$$\left|\frac{f^{\prime }(z)}{1-|f^2(z)}\right|\le \left|\frac{1}{1-|z{|}^2}\right|$$

6. PG 136, Problem 2

Consider the map $g:\mathcal{H}\to \mathcal{D}$, where $\mathcal{D}$ is an open unit disk and $\mathcal{H}$ is the upper half plane, given by $g(z)=\frac{z-z_0}{z-\overline{z_0}}$, where $z_0\in \mathcal{H}$ is a fixed point. Consider the map $h(z)=\frac{z-f(z_0)}{z-\overline{f(z_0)}}$

And consider the map $h\circ f\circ g^{-1}:\mathcal{D}\to \bar{\mathcal{D}}$ This satisfies the conditions that the function is analytic on $\mathcal{D}$, and $|h(f(g^{-1}(z)))|\le 1$ for all $z\in \mathcal{D}$ also $h(f(g^{-1}(0)))=0$.

This gives $|h(f(g^{-1}(z)))|\le |z|⟹|h(f(z))|\le |g(z)|⟹\left|\frac{f(z)-f(z_0)}{f(z)-\overline{f(z_0)}}\right|\le \frac{z-z_0}{z-{\bar{z}}_0}$ taking $z\to z_0$ we get

$$\frac{|f^{\prime }(z_0)|}{Im(f(z_0))}\le \frac{1}{Im(z_0)}$$

7. PG 136, Problem 3

In problem 5, equality holds iff

$$\left|\frac{f(z)-w_0}{1-{\overline{w}}_{0}f(z)}\right|=\left|\frac{z-z_0}{1-{\overline{z}}_{0}z}\right|$$

which holds iff equality holds in $(36)$ with $M=R=1$ hence $|Sf(T^{-1}\zeta )|=|\zeta |$ where $S$ adn $T$ are linear fractional transformations.

This gives $Sf{T}^{-1}(\zeta )=c\zeta$ for some $c$. This gives $f=S^{-1}(cT(\zeta ))$ which is a compostition of $3$ linear fractional transformationsm, and hence is a Linear fractional transformation.

In Problem 6, equality holds iff $|h(f(g^{-1}(z)))|=|z|$ which gives $h(f(g^{-1}(z)))=cz$ for some constant $c$. That gives $f$ as $f(z)=h^{-1}(cg(z))$ which is also a linear fractional transformation.

8. PG 148, Problem 2

Let $S$ be a simply connected space. Let $P=S\setminus M$ where $M$ is a set with $m$ points The $P^c=S^C\cup M$ points which are $m+1$ connected components, hence $p$ has connectivity $m+1$

The homology basis consists of $m$ loops each centered around the $m$ points. So for each of the $m$ points, we can find a circle ${\gamma }_i$ around the points $a_i\in M$ and we will have $n({\gamma }_i,a_j)={\delta }_{i,j}$ which forms a basis

9. PG 148, Problem 5

Any closed curve $\gamma$ in the domain $\Omega$ that winds around $1$ also winds around $-1$. consider some point $z_0\in \gamma$ and choose some value of ${\theta }_1=arg(1-z)$ such that $(1-z_0)=r_1{e}^{i{\theta }_1}$ . and take a branch of $\sqrt{1-z_0}$. Now as we travel around $\gamma$, as we come back to $z_0$ we add $2\pi$ to the argument of $(1-z)$ . The same works for the argument of $(1+z)$ and so the argument of the product changes by $4\pi$. Now when we take the square root, we divide the argument by $2$ which means that as we move around $\gamma$ and come back to $z_0$, the argument of $\sqrt{1-z^2}$ changes by $2\pi$.

Now we compute

$${\int }_{\gamma }\frac{1}{\sqrt{1-z^2}}dz.$$

And by Cauchy’s Theorem, we can assume $\gamma$ is a very large cirle and so acts as a small disc about infinity. we apply the change of variables $z=\frac{1}{w}$ and we get the integral

$$-{\int }_{|w|=\delta }\frac{dw}{w\sqrt{w^2-1}}$$

$\sqrt{w^2-1}$ is analytic in a neighbourhood of $0$ so we can compute the integral by cauchy’s formula to be $2\pi$