Answer to Question #235246 in Complex Analysis for Fozia Sayda

Question

Evaluate the integral $\oint$ e^zt/z²+1 dz where c is the circle |z|=3.

Answer 1

Inside the contour $C=\{z\in\mathbb{C}:|z|=3\}$, the function $\frac{e^{zt}}{z^2+1}$ has 2 simple poles at $z=i$ and $z=-i$. Hence, according to the residue theorem,

$\oint\limits_C \frac{e^{zt}}{z^2+1}dz=2\pi i({\rm res}_{z=i}\frac{e^{zt}}{z^2+1}+{\rm res}_{z=-i}\frac{e^{zt}}{z^2+1})$

$\operatorname{res}_{z=i}\frac{e^{zt}}{z^2+1}=\lim\limits_{z\to i}\frac{e^{zt}}{z^2+1}(z-i)=\lim\limits_{z\to i}\frac{e^{zt}}{z +i}=\frac{e^{it}}{2i}$

${\rm res}_{z=-i}\frac{e^{zt}}{z^2+1}=\lim\limits_{z\to i}\frac{e^{zt}}{z^2+1}(z+i)=\lim\limits_{z\to i}\frac{e^{zt}}{z -i}=-\frac{e^{-it}}{2i}$

Therefore,

$\oint\limits_C \frac{e^{zt}}{z^2+1}dz=2\pi i(\frac{e^{it}}{2i}-\frac{e^{-it}}{2i})=2\pi i\sin t$

Answer. $\oint\limits_C \frac{e^{zt}}{z^2+1}dz=2\pi i\sin t$