数学代写|MTH829 Complex Analysis

What conditions on $a$ and $b$ guarantee that $z^a(1-z)^b$ can be defined as a single-valued function on $\mathbb{C} \backslash[0,1]$ ? In this case describe the Riemann surface of this function?

Show that the set of points $\left{(z, w) \in \mathbb{C}^2 \mid w^2=\sin z\right}$ is a Riemann surface.

Let $P(z)$ be a polynomial with $2 n+1$ distinct complex roots. Show that $X=$ $\left{(z, w) \mid w^2=P(z)\right}$ is a Riemann surface. What is its genus?

With the notation from the lecture notes, show that for every $\omega_1$ and $\omega_2$ such that $\operatorname{Im} \omega_2 / \omega_1>0$, there is a $\tau$ in the upper half plane such $\mathbb{C} / \Lambda$ and $\mathbb{C} / L$ are conformally equivalent.

Note that $z_n$ is a double pole of $1 / \cos ^2 z$ for any integer $n$. Since $f(z)$ is analytic on the whole real axis, so $f(z)=\sum_{k=0}^{\infty} a_k\left(z-z_n\right)^k$, where $a_k=\frac{f^{(k)}\left(z_n\right)}{k !}$.
Also, $\frac{1}{\cos ^2 z}=\frac{b_2}{\left(z-z_n\right)^2}+\frac{b_1}{\left(z-z_n\right)}+\sum_{k=0}^{\infty} c_k\left(z-z_n\right)^k$ (since $z_n$ is a double pole). As shown in later calculations, it is necessary to find the value of $b_2$. The trick is to observe that $\frac{z-z_n}{\cos ^2 z}$ has a simple pole at $z_n$ and $\operatorname{Res}\left(\frac{z-z_n}{\cos ^2 z}, z_n\right)=b_2$. Note that
$$
b_2=\lim {z \rightarrow z_n} \frac{\left(z-z_n\right)^2}{\cos ^2 z}=\lim {z \rightarrow z_n} \frac{2\left(z-z_n\right)}{-2 \sin z \cos z}=\lim {z \rightarrow z_n} \frac{2}{-2 \cos ^2 z+2 \sin ^2 z}=1 \text {. } $$ Furthermore, since the Taylor series of $\cos ^2 z$ at $z=z_n$ has even power terms only (see below), the even power terms in the Laurent expansion of $\frac{1}{\cos ^2 z}$ at $z=z_n$ are zero. In particular, we have $b_1=0$. Consider the following series expansion $$ \frac{f(z)}{\cos ^2 z}=\left(\sum{k=0}^{\infty} a_k\left(z-z_n\right)^k\right)\left(\frac{b_2}{\left(z-z_n\right)^2}+\sum_{k=0}^{\infty} c_{2 k}\left(z-z_n\right)^{2 k}\right)
$$
so that the coefficient of $\frac{1}{z-z_n}$ is $a_1 b_2$. Hence, we obtain
$$
\operatorname{Res}\left(\frac{f(z)}{\cos ^2 z}, z_n\right)=a_1 b_2=a_1=f^{\prime}\left(z_n\right)
$$

Alternative method (if $f\left(z_n\right) \neq 0$ )
$$
\begin{aligned}
\cos ^2 z & =2\left(z-z_n\right)^2-8\left(z-z_n\right)^4+32\left(z-z_n\right)^6-\cdots \
& =\left(z-z_n\right)^2\left{2-8\left(z-z_n\right)^2+32\left(z-z_n\right)^4-\cdots\right}
\end{aligned}
$$
so $z_n$ is a double pole of $\frac{1}{\cos ^2 z}$. Using the formula in Example 6.2.2 (p. 230), we have
$$
\operatorname{Res}\left(\frac{f(z)}{\cos ^2 z}, z_n\right)=\left.\frac{6 f^{\prime}(z)\left(\cos ^2 z\right)^{\prime \prime}-2 f(z)\left(\cos ^2 z\right)^{\prime \prime \prime}}{3\left[\left(\cos ^2 z\right)^{\prime \prime}\right]^2}\right|_{z=z_n}=f^{\prime}\left(z_n\right) .
$$