These are just exercises from this list (The file is likely to disappear or change someday so if that happens I'll remove the link, too).
Exercise 1. The analytic function
$$f(z)=\frac{z e^{\sin(z+1)}}{(z-(4+2i)(z^2-19i))}$$
has a power series around $z=0$. Find the radius of convergence ot this series.
Solution. The radius of convergence of this series is just $\min |z_0|$ where $z_0$ ranges over the nonremovable singularities of $f$. These singularities are just where $z=4+2i$ and $z^2=19i$ (since $ze^{\sin(z+1)}\neq 0$ for those z\neq 0). In the first case, $|z|=2\sqrt{5}=\sqrt{20}$. In the second case, since $|z^2|=|z|^2=19$ then $|z|=\sqrt{19}$. Since $\sqrt{19}<\sqrt{20}$ then the radius of convergence of this series is just $\sqrt{19}$.
Exercise 2. Let $f$ be a meromorphic, periodic function: $f(z+c)=f(z)$. Prove that for all $z_0$, $\Res_{z_0}(f)=\Res_{z_0+c}(f)$.
Proof. Take a circle $C$ small enough around $z_0$ such that $f$ is holomorphic in $\text{Int}C-\{z_0\}$. Then the function $f(z+c)$ is holomorphic in $\text{Int}(C+c)-\{z_0+c\}$ and
$$\Res_{z_0}(f)=\frac{1}{2\pi i}\int_C f(z)dz=\frac{1}{2\pi i}\int_C f(z+c)dz=\frac{1}{2\pi i}\int_{C+c} f(z)dz=\Res_{z_0+c}(f).$$
Exercise 3. Let $g(z)=f(az)$. Compute $\Res_{z_0}(g)$ in terms of $f$.
Solution. $z_0$ is an isolated singularity of $g$, iff $az_0$ is an isolated singularity of $f$. So if
$$f(z)=\sum_{n\in \bZ} a_n (z-az_0)^n$$
in a neighborhood of $az_0$, then
$$g(z)=f(az)=\sum_{n\in\bZ} a_n(az-az_0)^n=\sum_{n\in\bZ} a^n a_n(z-z_0)^n$$
in a neighborhood of $z_0$. For $n=-1$, this gives the coefficient $\Res_{z_0} g=\frac{a_{-1}}{a}=\frac{\Res_{az_0} f}{a}$.
Exercise 4. Let $f$ be a holomorphic function with a triple zero at $z_0$. Compute $\Res_{z_0} \frac{1}{f}$ in terms of $f$.
Solution. Since $z_0$ is a triple zero, $f'(z_0)=f''(z_0)=0$, which means that a series expansion of $f$ around $z_0$ will be of the form
$$\sum_{n=3}^\infty a_n (z-z_0)^n.$$
The function $\frac{1}{f}$ will have a triple pole at $z_0$, so a series expansion around $z_0$ will have the form
$$\sum_{n=-3}^\infty b_n (z-z_0)^n.$$
Since $\frac{1}{f}f=1$, and multiplying the series we get $a_3 b_{-3}=1$ from which $b_{-3}=\frac{1}{a_3}$.
We also get $a_4 b_{-3}+a_3 b_{-2}=\frac{a_4}{a_3}+a_3 b_{-2}=0$, and $\frac{a_5}{a_{3}}+a_4 b_{-2}+a_3 b_{-1}=0$.
So
$$b_{-2}=-\frac{a_4}{a_3^2}$$
and
$$b_{-1}=-\frac{\frac{a_5}{a_3}+a_4 b_{-2}}{a_3}=-\frac{\frac{a_5}{a_3}- \frac{a_4^2}{a_3^2}}{a_3}.$$
But $b_{-1}$ is the residue and $a_n=\frac{f^{(n)}}{n!}$, so we're done.
Exercise 5 is about calculating some integrals.
The first one is
$$I_1=\int_{\partial B_1(0)} \frac{e^{iz}}{z^2}dz.$$
Since $e^{iz}=1+iz+\cdots$, then $\frac{e^{iz}}{z}=\frac{1}{z^2}+\frac{i}{z}+\cdots$, so by the residue theorem $I_1=-2\pi$.
The next one is
$$\int_{\partial B_r(a)}\frac{dz}{z-a}$$
which by a Riemann sum argument is known to be $2\pi i$.
The next one is $I_3=\int_{\partial B_1(0) }\frac{\sin z}{z^3}dz$. Since all the even coefficients in the Maclaurin series of $\sin z$ are $0$, then all the odd coefficients of the Laurent series of $ \frac{\sin z}{z^3}$ are $0$, in particular the coefficient of $z^{-1}$. Thus, $I_3=0$.
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