I am trying to calculate $\int_C{\frac{e^{2\pi z}}{(z^2+4)^2}} dz$, where $C$ is the ellipse given by the equation $196883x^2 + (y-2)^2 = 1$ oriented positively.
So my first reflex was to check if the function we are integrating was analytic over the domain enclosed by $C$. By checking the roots of $(z^2 + 4)$, I get that $z=\pm2i$. Clearly, $z=2i$ is in the domain, therefore I tried to define $B(z_0,r)$ an open ball centered at $z_0 = 2i$ and of radius $r > 0$ for a small $r$. Now let $\gamma$ be the positive oriented closed curve that enclosed $B(z_0,r)$.Then by Cauchy I get the following equality : $\int_C{\frac{e^{2\pi z}}{(z^2+4)^2}} dz - \int_\gamma{\frac{e^{2\pi z}}{(z^2+4)^2}}dz = 0$. So I am now trying to compute that second integral, but after using a change of variable with $z=e^{it}, dz= ie^{it}dt$ for $t\in [0, 2\pi]$, I get an integral that I don't know how to solve. Is the beginning of my reasoning right ? And if yes, any help to compute that last integral would be more than welcomed.
numerator=$e^{2\pi ((z-2i)+2i)}=e^{4\pi i }e^{2 \pi (z-2i)}=e^{2 \pi (z-2i)}$ $$\text {denominator=}((z-2i)+4i)^2(z-2i)^2$$ $$=-16(1-(i/4)(z-2i))^2(z-2i)^2$$ Now you an write the whole expression as a Laurent series in $z-2i$ and thus find the residue at $2i.$ Note that $2i$ is the only pole inside the given curve.