Evaluate $\int_{C}\frac{f(z)}{z^{3}}\,dz$

Question

Let $f(z)=\displaystyle\sum_{k=0}^{\infty}{k^{4}\left(\dfrac{z}{4}\right)^{k}}$. Evaluate $\displaystyle\int_{C}{\dfrac{f(z)}{z^{3}}dz}$, where $C$ is the circle $\vert z\vert=\pi$ positively oriented.

My approach: Let $z=e^{i\theta}\pi$ with $0\leq \theta\leq 2\pi$, then $\vert z\vert=\pi$. And also, $dz=ie^{i\theta}\pi d\theta=izd\theta\to \dfrac{dz}{z}=id\theta$. So

$$\displaystyle\int_{C}{\dfrac{f(z)}{z^{3}}dz}=\int_{0}^{2\pi}{\dfrac{f(e^{i\theta}\pi)}{(e^{i\theta}\pi)^{2}}e^{i\theta}\pi id\theta}=i\int_{0}^{2\pi}{\dfrac{f(e^{i\theta}\pi)}{(e^{i\theta}\pi)^{2}}d\theta}$$

But, $f(e^{i\theta}\pi)=\displaystyle\sum_{k=0}^{\infty}{k^{4}\left(\dfrac{e^{i\theta}\pi}{4}\right)^{k}}$, then $$\int_{C}{\dfrac{f(z)}{z^{3}}dz}=i\int_{0}^{2\pi}{\sum_{k=0}^{\infty}{\dfrac{k^{4}(e^{i\theta}\pi)^{k-2}}{4^{k}}}d\theta}$$ But I don't how continuous this, regards!

Answer 1

By the form it has, we know that $f(z)$ is analytic. Then we can apply the Cauchy Integral Formula:

$$\int_C\frac{f(z)}{z^3}dz=\pi i f''(0).$$

Taking derivatives of $f$ we get: $$f'(z)=\sum_{k=1}^\infty\frac{k^5}{4}\left(\frac{z}{4}\right)^{k-1}.$$ $$f''(z)=\sum_{k=2}^\infty\frac{k^5(k-1)}{16}\left(\frac{z}{4}\right)^{k-2}.$$ Evaluating at $0$, we get $f''(0)=2$, and the integral has a value of $2\pi i$.

Answer 2

$$I=i\int_{0}^{2\pi}{\sum_{k=0}^{\infty}{\dfrac{k^{4}(e^{i\theta}\pi)^{k-2}}{4^{k}}}d\theta} =i {\sum_{k=0}^{\infty}{\dfrac{k^{4}(\pi)^{k-2}}{4^{k}}}} \int_{0}^{2\pi}e^{i(k-2)\theta}d\theta$$

but $\int_{0}^{2\pi}e^{i(k-2)\theta}d\theta$ vanishes unless $k=2$ when it takes the value $ 2 \pi $

So

$$ I = 2 \pi i {\dfrac{2^{4}(\pi)^{0}}{4^{2}}} = 2 \pi i$$