数学代写|MTH829 Complex Analysis

(a) (Qual Aug 1996) Suppose $f(z)=\sum_{n=0}^{\infty} a_n(z-c)^n$ has the property that the series $\sum_{n=0}^{\infty} f^{(n)}(c)$ converges. Show that $f$ is an entire function (i.e. a function that can be expressed as power series with radius of convergence $R=\infty$ ).
(b) What if the convergence of $\sum_{n=0}^{\infty} f^{(n)}(c)$ is replaced by the assumpt ion that the power series
$$
\sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{(n+1)^{2002}} z^n
$$
has a positive radius of convergence?

(HE Chapter 2, ex. 30) (Qual Aug 1996) Show that an entire function that takes real values on the real axis and purely imaginary values on the imaginary axis must be an odd function:
$$
f(-z)=-f(z), \quad \text { for all } z \in C .
$$

Compute the radius of convergence of the following series:
(a) $\sum_{n=1}^{\infty} \frac{(2 n) !}{n^{2 n}} z^n$ (Qual Aug 1994).
(b) $\sum_{n=0}^{\infty}\left(n+a^n\right) z^n$, where $a \in C$ (Qual Aug 1994).
(c) The Taylor series around zero for the function $z \cot z$ (Qual Aug 1994).
(d) $\sum_{n=1}^{\infty} \frac{(3 n) !}{(3 n)^{3 n}} z^n$ (Qual Jan 1994).
(e) $\sum_{n=1}^{\infty}\left(3+(-1)^n\right)^n z^n$ (Qual Jan 1994)
(c) The Taylor series around zero for the function $\frac{z}{e^z-1}$ (Qual Jan 1994).

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
$$
b_2=\lim {z \rightarrow z_n} \frac{\left(z-z_n\right)^2}{\cos ^2 z}=\lim {z \rightarrow z_n} \frac{2\left(z-z_n\right)}{-2 \sin z \cos z}=\lim _{z \rightarrow z_n} \frac{2}{-2 \cos ^2 z+2 \sin ^2 z}=1 .
$$

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
$$
\operatorname{Res}\left(\frac{f(z)}{\cos ^2 z}, z_n\right)=\left.\frac{6 f^{\prime}(z)\left(\cos ^2 z\right)^{\prime \prime}-2 f(z)\left(\cos ^2 z\right)^{\prime \prime \prime}}{3\left[\left(\cos ^2 z\right)^{\prime \prime}\right]^2}\right|_{z=z_n}=f^{\prime}\left(z_n\right)
$$