Can someone please help me answer this question as I cannot seem to get to the answer. Please note that the Cauchy integral formula must be used in order to solve it.
Many thanks in advance! \begin{equation*} \int_{|z|=3}\frac{e^{zt}}{z^2+4}=\pi i\sin(2t). \end{equation*}
Also $|z| = 3$ is given the counterclockwise direction.
You should know that using the residue theorem would be easier, but if we are to restrict ourselves to Cauchy's integral formula, then here's one way of attacking it:
First, note that the integrand is a quotient of two entire functions. As such, the integrand is analytic everywhere except the points at which the denominator is zero. Since $z^2+4$ can be factored as $(z-2i)(z+2i)$, then the only points at which the integrand is not analytic are $\pm 2i$. Unfortunately, both of these points are inside the circle $|z| = 3$, so in order to apply Cauchy's integral formula, we will have to be clever.
Let $C$ be the circle $|z| = 3$ oriented counterclockwise. Let $C_1$ be the upper half of the circle $|z| = 3$ together with the line segment $[-3, 3]$ oriented counterclockwise. Let $C_2$ be the lower half together with the line segment $[-3, 3]$ oriented counterclockwise. Notice we have:
$$\int_C \frac{e^{zt}}{z^2+4} \ dz = \int_{C_1} \frac{e^{zt}}{z^2+4} \ dz + \int_{C_2} \frac{e^{zt}}{z^2+4} \ dz$$
Now we can attack the two integrals on the right hand side separately using Cauchy's integral formula. For the first, e.g., you can let $\displaystyle f(z) = \frac{e^{zt}}{z+2i}$, which is analytic everywhere inside $C_1$, and your integrand becomes $\displaystyle \frac{f(z)}{z-2i}$.
Alternative method:
Using partial fraction decomposition, we have $\displaystyle \frac{1}{z^2+4} = \frac{i}{4(z+2i)} - \frac{i}{4(z-2i)}$. Hence:
$$\int_C \frac{1}{z^2+4} \ dz = \int_C \frac{i}{4(z+2i)} \ dz - \int_C \frac{i}{4(z-2i)} \ dz$$
And then one can apply Cauchy's integral formula on the two separate pieces without having to split the contour.
Write integral in the form the function in the form $\frac{e^{zt}}{(z-2i)(z+2i)}$. Then you should split the contour in two parts such that interior of each part contain only one point $2i$ or $-2i$.
Then apply Cauchy integral formula for each contour.
Since all the poles of the integrand are enclosed by the contour, we have \begin{eqnarray} \int_{|z|=3}\frac{e^{zt}}{z^2+4}dz=2\pi i\text{Res}_{z=0}\frac{1}{z^2}\frac{e^{\frac{t}{z}}}{\frac{1}{z^2}+4}=2\pi i\text{Res}_{z=0}\frac{e^{\frac{t}{z}}}{1+4z^2} \end{eqnarray} Note that $\displaystyle \frac{e^{\frac{t}{z}}}{1+4z^2}=\left(\sum_{n=0}^\infty (-1)^n4^nz^{2n}\right)\left(\sum_{n=0}^\infty \frac{t^n}{n! z^n}\right)$. So $\displaystyle\text{Res}_{z=0}=\sum_{n=0}^\infty \frac{(-1)^n4^nt^{2n+1}}{(2n+1)!}=\frac{1}{2}\sin 2t$
Hint. The denominator $\displaystyle z^2+4=(z-2i)(z+2i)$ of the function $$ z \longmapsto \frac{e^{zt}}{z^2+4} $$ gives two poles inside $|z|=3$ and by the Cauchy integral formula we have $$ \int_{|z|=3}\frac{e^{zt}}{z^2+4}dz=2i\pi\left({\rm{Res}}_{z=-2i}f(z)+{\rm{Res}}_{z=2i}f(z)\right). $$ You conclude using for example $$ {\rm{Res}}_{z=-2i}f(z)=\lim_{z\to -2i}\left((z+2i)\times\frac{e^{zt}}{z^2+4}\right)=\ldots $$
