I am calculating the residue of $$ \frac{\sin z}{z^{2}(\pi-z)} $$
Since there is only $1$ pole (simple) at $0$, I went about calculating the residue according to the rule: $$Res\left [ \frac{f(x)}{g(x)}, z_{0} \right ] = \lim_{z\rightarrow z_{0}} \frac{f(x)}{g'(x)}$$
With this formula, I find the residue to be $\frac{1}{2\pi}$. However, the answer is supposed to be $\frac{1}{\pi}$.
Is the formula I have used correct? Or am I making some other mistake?
The Residual at a single pole is the exact value of the function $f(z)$, multiplied by the pole factor, at the pole value. So here, near z =0:
$$ f(z) = \frac{\sin z}{z^{2}(\pi-z)} = \frac{1}{z (\pi-z)} + \cal O (z) $$
So
$$ \mathrm {Res} f(z) |_{z=0} = \lim_{z \to 0} {\Large [} z\cdot f(z) {\Large ]} = \lim_{z \to 0} {\Large [} \frac{1}{(\pi-z)} + \cal O (z^2) {\Large ]} = \frac{1}{\pi} $$
If $f$ has a simple pole in $z_0$ then
$Res(f;z_0)= \lim_{z \to z_0}(z-z_0)f(z)$. Hence , in your case:
$Res(f;0)= \lim_{z \to }\frac{\sin }{z}* \frac{1}{\pi-z}=1*\frac{1}{\pi}=\frac{1}{\pi}$
