Evaluating the integral $\frac{1}{2\pi i} \oint_C\frac{e^{zt}}{z^2+1}dz$ with Cauchy's theorem.

Question

I would like to show that

$$\frac{1}{2\pi i} \oint_C\frac{e^{zt}}{z^2+1}dz=\sin t$$ if $t>0$ and $C$ is the circle $|z|=3$.

I am pretty sure that I need to use Cauchy to do this because it is really similar to the format, $$f(a)=\frac{1}{2\pi i} \oint_C\frac{f(z)}{z-a}dz$$ Only, the $z^2$ is throwing me off. Can I break this up into multiple integrals somehow?

Answer 1

Hint: Note that $$\dfrac{1}{z^2+1}=\dfrac{1}{-2i}\left(\dfrac{1}{z+i}-\dfrac{1}{z-i}\right)$$

Edit: \begin{align} \frac{1}{2\pi i}\oint_C\frac{e^{zt}}{z^2+1}\ dz &= \frac{1}{2\pi i}\oint_C\dfrac{1}{-2i}\dfrac{e^{zt}}{z+i}\ dz - \frac{1}{2\pi i}\oint_C\dfrac{1}{-2i}\dfrac{e^{zt}}{z-i}\ dz \\ &= \dfrac{1}{-2i}\left(e^{-it}-e^{it}\right) \\ &= \sin t \end{align}