Compute $\int_{|z|=1}\overline z^2$ $\sin z$ $\mathrm dz$

Question

The solution provided made use of the taylor expansion of $\sin z$ which i thought was quite tedious.

My solution:

On $|z|=1$, $\overline z=\frac{1}{z} $ Hence, $\int_{|z|=1} f(z) \,\mathrm dz$ = $\int_{|z|=1}(\overline z)^2\sin z \,\mathrm dz$ = $\int_{|z|=1} \frac{\sin z}{z^2}\,\mathrm dz$.

The second integral can be solved using Riemann integral formula which gives 2$\pi$i which is also the answer. Is my solution correct?

Answer 1

Just by applying Cauchy's integral formula $$f'(0)=\frac{1}{2\pi i}\int\limits_{|z|=1} \frac{f(z)}{z^2}dz$$ where $f(z)=\sin{z} \Rightarrow f'(z)=\cos{z} \Rightarrow f'(0)=1$.

Answer 2

The residue theorem can be used since $|z|=1$ encloses $z=0$, where $f(z)$ has a pole of order $1$

$$\operatorname{Res}(f,0)=\frac{1}{0!}\lim_{z\to0}\frac{\sin(z)}{z}=1$$

Then

$$\oint_\Gamma\frac{\sin z}{z^2}=2\pi i\sum_{j=1}^n\operatorname{Res}(f,z_j)=2\pi i$$