Evaluating the following integral by my professor, we have:
$$\int_0^{\frac{\pi}{2}} \sin^4 \theta = \frac{1}{4}\int_0^{2 \pi} \sin^4 \theta= \frac{1}{4}\int_0^{2 \pi}(\frac{e^{i\theta}-e^{-i\theta}}{2i})^4 \frac{e^{i\theta}}{e^{i\theta}}=\frac{1}{4}\int_{|z|=1}(\frac{z-\frac{1}{z}}{2i})^4\frac{1}{iz}dz\\ =\frac{1}{64i}\int_{|z|=1}\frac{1}{z}(z^4-4z^2+6-4\frac1{z^2}+\frac1{z^4})dz= \frac{1}{64i}2\pi i \cdot 6$$
Here, $6 =res (\frac{1}{z}(z^4-4z^2+6-4\frac1{z^2}+\frac1{z^4};1)$
Why am I allowed to use the residue theorem at the singularity $1$, despite $1$ is on the boundary of my integration path $|z|=1$ ? I thought the closed curve isn't allowed to meet any of $a_k$.
There is no singularity at $z=1$. The only singularity is at $z=0$ so Residue Theorem can be applied.