Finding a Laurent series that converges for $\frac{sin(z)}{(z-{\pi}/4)^3} $:
What I have done:
Figured out that the Laurent Series can be expressed as this:
$$\sum_{n=0}^{\infty} \frac{\sqrt{2}*[cos \frac{n\pi}{2} +sin\frac{n\pi}{2}]*(z-\frac{\pi}{4})^{n-3}}{2*(n!)}$$
However when I attempt to do the ratio test, I am stuck at:
$$\lvert \frac{[cos \frac{(n+1)\pi}{2} +sin\frac{(n+1)\pi}{2}]*(z-\frac{\pi}{4})}{(n+1)(cos \frac{n\pi}{2} +sin\frac{n\pi}{2})}\rvert$$
Does anyone have any good guidance as to how I should proceed forward?
Would it be more beneficial to express cosine and sine function differently? All advice is appreciated.
Edit: If you wish to know how I managed to express the series in that way, please let me know, thank you.
Taking from what you have there, notice that $\cos(n\pi/2) + \sin(n\pi/2)$ takes on either $-1$ or $1$. The limit then simplifies to
$$\lim_{n\to \infty}\left| \frac{z-\pi/4}{n+1}\right|,$$
which converges to 0.