Need help getting started on this Complex Integral.

Question

Here is the problem statement:

Find $\displaystyle \oint_{_{C}} z^{3}\cos{\left(\frac{3}{z}\right)}dz$ for $C=|z|=1$

recall that $\displaystyle \cos{(z)}=\sum\limits_{n=0}^{\infty}\frac{(-1)^{n}}{(2n)!}z^{2n}$

I'm not sure how to get started on this problem. There is an issue with $z=0$ due to the cos function but I'm not sure how to implement the series form into the integral and then perform the integral. Any help would be greatly appreciated.

EDIT Using the comment by @Conrad below.

For $\displaystyle \oint_{_{C}}z^{3}\cos{\left(\frac{3}{z}\right)}dz$

Let $\displaystyle w=\frac{1}{z}\Rightarrow -dw=\frac{1}{z^2}dz$

Then rewrite the integral using a factor of 1 written as $\displaystyle 1=\frac{z^2}{z^2}$

$\displaystyle \Rightarrow \oint_{_{C}}\frac{z^{3}z^{2}\cos{\left(\frac{3}{z}\right)}}{z^2}dz$, next we swap out the $z$ and $dz$ and arrive at

$\displaystyle \Rightarrow \oint_{_{C}}\frac{-\cos{(3w)}}{w^{5}}dw$, here we have a pole of order $5$ at $w=0$ and use the Residue Theorem to finish out.

Result using the residue theorem as stated below $\displaystyle \oint_{_{C}}\frac{-\cos{(3w)}}{w^{5}}dw = 2\pi i\left[\frac{1}{(n-1)!}\lim\limits_{w\to z_{0}}\frac{d^{n-1}}{dw^{n-1}}(w-z_{0})^{n} \frac{-\cos{(3w)}}{w^{5}} \right]$

plugging in $n=5$ and $z_{0}=0$

$\displaystyle \oint_{_{C}}\frac{-\cos{(3w)}}{w^{5}}dw = 2\pi i\left[\frac{1}{(5-1)!}\lim\limits_{w \to 0}\frac{d^{5-1}}{dw^{5-1}}\left((w-0)^{5}\frac{-\cos{(3w)}}{w^{5}}\right) \right]$

$\displaystyle \oint_{_{C}}\frac{-\cos{3(w)}}{w^{5}}dw = 2\pi i\left[\frac{1}{24}\lim\limits_{w \to 0}\frac{d^{4}}{dw^{4}}\left(-\cos{(3w)}\right) \right]$

$\displaystyle \oint_{_{C}}\frac{-\cos{(3w)}}{w^{5}}dw = 2\pi i\left[\frac{1}{24}\lim\limits_{w \to 0}\left(-81\cos{(3w)}\right) \right]$

$\displaystyle \oint_{_{C}}\frac{-\cos{(3w)}}{w^{5}}dw = 2\pi i\left[\frac{1}{24}\cdot(-81) \right]$

$\displaystyle \oint_{_{C}}\frac{-\cos{(3w)}}{w^{5}}dw = -\frac{27\pi i}{4}$ or $\displaystyle \frac{27\pi}{4i}$

I am trying to figure out why my answer is off by a factor of -1 to make it align with the correct answer @bob provided below.

Answer 1

We have the Laurent series, $$f(z)=z^3\cos\left(\frac{3}{z}\right)=\sum_{n=0}^\infty\frac{(-9)^n}{(2n)!}z^{3-2n}$$ as $z\to0$, so by Cauchy's residue theorem, $$\oint_{|z|=1}f(z)\ dz=2\pi i\text{Res}(f,0)=2\pi i\cdot\frac{81}{4!}=\frac{27\pi i}{4}.$$

Answer 2

EDIT: I made mistakes in this one and just updated the original post

Using the comment by @Conrad above.

For $\displaystyle \oint_{_{C}}z^{3}\cos{\left(\frac{3}{z}\right)}dz$

Let $\displaystyle w=\frac{1}{z}\Rightarrow -dw=\frac{1}{z^2}dz$

Then rewrite the integral using a factor of 1 written as $\displaystyle 1=\frac{z^2}{z^2}$

$\displaystyle \Rightarrow \oint_{_{C}}\frac{z^{3}z^{2}\cos{\left(\frac{3}{z}\right)}}{z^2}dz$, next we swap out the $z$ and $dz$ and arrive at

$\displaystyle \Rightarrow \oint_{_{C}}\frac{-\cos{(w)}}{w^{5}}dw$, here we have a pole of order $5$ at $w=0$ and use the Residue Theorem to finish out.

Result using the residue theorem as stated below $\displaystyle \oint_{_{C}}\frac{-\cos{(w)}}{w^{5}}dw = 2\pi i\left[\frac{1}{(n-1)!}\lim\limits_{w\to z_{0}}\frac{d^{n-1}}{dw^{n-1}}(w-z_{0})^{n} \frac{-\cos{(w)}}{w^{5}} \right]$

plugging in $n=5$ and $z_{0}=0$

$\displaystyle \oint_{_{C}}\frac{-\cos{(w)}}{w^{5}}dw = 2\pi i\left[\frac{1}{(5-1)!}\lim\limits_{w \to 0}\frac{d^{5-1}}{dw^{5-1}}\left((w-0)^{5}\frac{-\cos{(w)}}{w^{5}}\right) \right]$

$\displaystyle \oint_{_{C}}\frac{-\cos{(w)}}{w^{5}}dw = 2\pi i\left[\frac{1}{24}\lim\limits_{w \to 0}\frac{d^{4}}{dw^{4}}\left(-\cos{(w)}\right) \right]$

$\displaystyle \oint_{_{C}}\frac{-\cos{(w)}}{w^{5}}dw = 2\pi i\left[\frac{1}{24}\lim\limits_{w \to 0}\left(-\cos{(w)}\right) \right]$

$\displaystyle \oint_{_{C}}\frac{-\cos{(w)}}{w^{5}}dw = 2\pi i\left[\frac{1}{24}\cdot(-1) \right]$

$\displaystyle \oint_{_{C}}\frac{-\cos{(w)}}{w^{5}}dw = -\frac{\pi i}{12}$ or $\displaystyle \frac{\pi}{12i}$