Proving that $\frac{1}{2\pi i}\int_{\gamma(0,3)}\frac{e^{zb}}{z^2+1}dz= \sin (b)$.

Question

So essentially I want to prove that $\frac{1}{2\pi i}\int_{\gamma(0,3)}\frac{e^{zb}}{z^2+1}dz= \sin (b), \forall b \in \mathbb{C}$.

I'm not entirely sure where to start but I had the following ideas to rewrite the equation: $\frac{1}{2\pi i}\int_{\gamma(0,3)}\frac{e^{zb}}{(z+1)(z-1)}dz= \sin(b)$

Then we can see we have a pole of order 1 at $z=1$ (I think). Potentially I could rewrite $e^{zb}= \cos(b)+i\sin(b)$. But I would have to assume that $b$ is of the form $a+ci$ where $a=0$. I dont know if I can really do this..

Also I think I could use the residue or perhaps the Cauchy Integral Formula but these are all just guesses.

Any points in the right direction would be greatly appreciated.

Answer 1

Note that $(z+1)(z-1)=z^2-1\neq z^2+1$.

I suppose that your path is $\gamma\colon[0,2\pi]\longrightarrow\mathbb C$ defined by $\gamma(t)=3e^{it}$. If so, apply the residue theorem and the fact that $z^2+1=(z+i)(z-i)$ in order to get that your integral is equal to$$\frac{e^{ib}}{2i}+\frac{e^{-ib}}{-2i}=\frac{e^{ib}-e^{-ib}}{2i}=\sin b.$$

This problem can also be solved using the Cauchy integral formula instead of the residue theorem.