Problem 1. Does there exist a function $f: \mathbb{C} \rightarrow \mathbb{C}$ that is holomorphic at every point on the unit circle $\mathbb{S}^{1}=\{z \in \mathbb{C}:|z|=1\}$ and not holomorphic anywhere else in the complex plane? If yes, provide such a function with complete justification. If not, explain why not.

Proof . Yes, such a function exists.
Consider the function $f(z)=(|z|-1)^{2}$. Then $f$ is infinitely real-differentiable for all $z \neq 0$. For such $z$
$$\frac{\partial f}{\partial \bar{z}}=2(|z|-1) \frac{z}{\bar{z}}$$
which is nonzero unless $|z|=1$. We have proved in class that a smooth function $f$ is holomorphic if and only if it satisfies the Cauchy-Riemann equations, namely $\partial f / \partial \bar{z}=0$. Therefore we conclude that $f$ is holomorphic at $z_{0} \in \mathbb{C} \backslash\{0\}$ if and only if $z_{0}$ satisfies $\left|z_{0}\right|=1$ Further if $z_{0}=0$, a direct computation shows that
$$\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{|h|^{2}-2|h|}{h}=-2 \lim _{h \rightarrow 0} \frac{|h|}{h}$$
does not exist, as can be seen by choosing sequences $h$ approaching 0 along the real and imaginary axis respectively. This proves that $f$ is not holomorphic at zero either, completing the proof.

Problem 2. For all possible values of $a_{0}$ and $a_{1}$, find the radius of convergence of the power series
$$\sum_{n=0}^{\infty} a_{n} z^{n}$$
where $a_{n}=a_{n-1}+a_{n-2}$ for all $n>1$.
Proof . Set $f(z)=\sum_{n=0}^{\infty} a_{n} z^{n}$. Clearly the radius of convergence $R=\infty$ if $a_{0}=a_{1}=0 .$ If at least one of $a_{0}$ and $a_{1}$ is nonzero, then the recursion relation above implies
$$\sum_{n=2}^{\infty} a_{n} z^{n}=\sum_{n=2}^{\infty} a_{n-1} z^{n}+\sum_{n=2}^{\infty} a_{n-2} z^{n} \quad \text { i.e., } f(z)-a_{0}-a_{1} z=z\left(f(z)-a_{0}\right)+z^{2} f(z)$$
Solving the equation leads to the following expression for $f$ :
(1)
$$f(z)=\frac{-a_{0}+\left(a_{0}-a_{1}\right) z}{z^{2}+z-1}$$
In other words $f$ is a rational function, which is holomorphic at all points $z$ except possibly where $z^{2}+z-1=0,$ i.e., $z=(-1 \pm \sqrt{5}) / 2 .$ Set
$$\alpha_{1}=\frac{-1+\sqrt{5}}{2}, \quad \alpha_{2}=\frac{-1-\sqrt{5}}{2}, \quad \text { so that }\left|\alpha_{1}\right|<\left|\alpha_{2}\right|$$
The radius of convergence of $f$ is determined by the possible cancellation of the numerator and denominator in (1). More precisely, two cases arise.

Case 1: either $a_{0}=a_{1}$ or $a_{0} /\left(a_{0}-a_{1}\right) \neq \alpha_{1} .$ In this case $f$ fails to be holomorphic at $\alpha_{1},$ and hence the power series expansion is valid on the smallest disc centred at the origin excluding $\alpha_{1} .$ In other words, $R=\left|\alpha_{1}\right|$.

Case 2: $a_{0} /\left(a_{0}-a_{1}\right)=\alpha_{1} .$ In this case, $f(z)=\left(a_{0}-a_{1}\right) /\left(z-\alpha_{2}\right),$ so its power series expansion is valid on $|z|<r=\left|\alpha_{2}\right|$.

=”” <=”” p=””></r=\left|\alpha_{2}\right|\).>

Problem 3. Express the integral $\int_{0}^{\pi} \frac{d \theta}{a+\sin ^{2} \theta} \quad$ in the form $\quad \oint_{|z|=1} \frac{f(z)}{z-z_{0}} d z$
and then use the Cauchy integral formula to evaluate it. Here $a>1$ is a fixed constant.

Proof . We write
\begin{aligned} \int_{0}^{\pi} \frac{d \theta}{a+\sin ^{2} \theta}=\int_{0}^{\pi}\left[a+\frac{1}{2}(1-\cos (2 \theta)]^{-1} d \theta\right.&=\frac{1}{2} \int_{0}^{2 \pi}\left[a+\frac{1}{2}(1-\cos \varphi]^{-1} d \varphi\right.\\ &=\int_{0}^{2 \pi}[(2 a+1)-\cos \varphi]^{-1} d \varphi \end{aligned}
where the second step uses the change of variable $\varphi=2 \theta .$ Set $z=e^{i \varphi} .$ Then, $d \varphi=\frac{1}{i} \frac{d z}{z},$ and
$$\cos \varphi=\frac{e^{i \varphi}+e^{-\varphi}}{2}=\frac{z+1 / z}{2}$$
Substituting these into the integral above, we obtain
$$\int_{0}^{\pi} \frac{d \theta}{a+\sin ^{2} \theta}=\frac{1}{i} \oint_{|z|=1} \frac{2 d z}{2(2 a+1) z-z^{2}-1}=-\frac{2}{i} \oint_{|z|=1} \frac{\frac{1}{z-z_{1}} d z}{z-z_{0}}$$
where $e$
$$\begin{array}{l} z_{0}=2 a+1-2 \sqrt{a^{2}+a}, \text { and } \\ z_{1}=2 a+1+2 \sqrt{a^{2}+a} \end{array}$$
are the two roots of the polynomial $z^{2}-2(2 a+1) z+1$. Note that $z_{1}$ lies outside the unit disk and $z_{0}$ lies in its interior. Therefore by the Cauchy integral formula with $f(z)=1 /\left(z-z_{1}\right)$, we obtain
$$\int_{0}^{\pi} \frac{d \theta}{a+\sin ^{2} \theta}=-\frac{2}{i} 2 \pi i \frac{1}{z_{0}-z_{1}}=\frac{\pi}{\sqrt{a^{2}+a}} .$$

Problem 4. Show that a conformal map preserves angles, in the following sense. If $f: U \rightarrow V$ is conformal, and $\Gamma_{1}, \Gamma_{2}$ are two curves in $U$ intersecting at $z_{0},$ then the angle between $\Gamma_{1}$ and $\Gamma_{2}$ at $z_{0}$ is the same as the angle between $f \circ \Gamma_{1}$ and $f \circ \Gamma_{2}$ at $f\left(z_{0}\right) .$ (Hint: Recall that the angle between two curves at an intersection point is, by definition, the angle between their tangents.)

Proof . Solution. Let $\gamma_{i}:[0,1] \rightarrow U$ be a parametrization of the curve $\Gamma_{i}$ with $\gamma_{i}\left(t_{i}\right)=z_{0} .$ The angle $\theta$ between $\Gamma_{1}$ and $\Gamma_{2}$ at $z_{0}$ is then the angle between the vectors $\gamma_{i}^{\prime}\left(t_{i}\right),$ hence
$$\theta=\arg \left(\gamma_{1}^{\prime}\left(t_{1}\right)\right)-\arg \left(\gamma_{2}^{\prime}\left(t_{2}\right)\right)$$
The corresponding angle between the image curves $f \circ \gamma_{i}$ at $f\left(z_{0}\right)$ is, by the same argument
\begin{aligned} \theta^{\prime} &=\arg \left(\left(f \circ \gamma_{1}\right)^{\prime}\left(t_{1}\right)\right)-\arg \left(\left(f \circ \gamma_{2}\right)^{\prime}\left(t_{2}\right)\right) \\ &=\arg \left(f^{\prime}\left(z_{0}\right) \gamma_{1}^{\prime}\left(t_{1}\right)\right)-\arg \left(f^{\prime}\left(z_{0}\right) \gamma_{2}^{\prime}\left(t_{2}\right)\right) \\ &=\arg \left(f^{\prime}\left(z_{0}\right)\right)+\arg \left(\gamma_{1}^{\prime}\left(t_{1}\right)\right)-\left[\arg \left(f^{\prime}\left(z_{0}\right)\right)+\arg \left(\gamma_{2}^{\prime}\left(t_{2}\right)\right)\right]=\theta \end{aligned}
Note that the second step follows from the chain rule, while the penultimate step uses the fact that $f^{\prime}\left(z_{0}\right) \neq 0$ (since $f$ is conformal), as a result of which $\arg \left(f^{\prime}\left(z_{0}\right)\right)$ is well-defined (even if multi-valued).

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