Laurent series of $\frac{sin(z)}{(z-{\Pi}/4)^3} $ in $0 < |z-\Pi/4| < 1$

Question

What will be Laurent series of $\frac{sin(z)}{(z-{\Pi}/4)^3} $ in $0 < |z-\Pi/4| < 1$ ?

I have calculated as ${z-\Pi/4} = t$ => $\frac{sin(t+\Pi/4)}{t^3}$ = $\frac{(t+\Pi/4) - (t+\Pi/4)^3/!3 + (t+\Pi/4)^5/!5......}{t^3}$

Solving and substituting $t = z - \Pi/4$ I get final answer. Is this the correct way to solve?

Answer 1

You'll probably want to avoid having series in the coefficients, thus use $$ \sin(t+\frac\pi4)=\frac{\sin t+\cos t}{\sqrt2}. $$