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.
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 \(n>1\).
$$
\sum_{n=0}^{\infty} a_{n} z^{n}
$$
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.
=”” <=”” p=””></r=\left|\alpha_{2}\right|\).>
$$
\sum_{n=2}^{\infty} a_{n} z^{n}=\sum_{n=2}^{\infty} a_{n-1} z^{n}+\sum_{n=2}^{\infty} a_{n-2} z^{n} \quad \text { i.e., } f(z)-a_{0}-a_{1} z=z\left(f(z)-a_{0}\right)+z^{2} f(z)
$$
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 \(|z|<r=\left|\alpha_{2}\right|\).
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.
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
$$
\int_{0}^{\pi} \frac{d \theta}{a+\sin ^{2} \theta}=-\frac{2}{i} 2 \pi i \frac{1}{z_{0}-z_{1}}=\frac{\pi}{\sqrt{a^{2}+a}} .
$$
$$
\begin{aligned}
\int_{0}^{\pi} \frac{d \theta}{a+\sin ^{2} \theta}=\int_{0}^{\pi}\left[a+\frac{1}{2}(1-\cos (2 \theta)]^{-1} d \theta\right.&=\frac{1}{2} \int_{0}^{2 \pi}\left[a+\frac{1}{2}(1-\cos \varphi]^{-1} d \varphi\right.\\
&=\int_{0}^{2 \pi}[(2 a+1)-\cos \varphi]^{-1} d \varphi
\end{aligned}
$$
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).
$$
\theta=\arg \left(\gamma_{1}^{\prime}\left(t_{1}\right)\right)-\arg \left(\gamma_{2}^{\prime}\left(t_{2}\right)\right)
$$
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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