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Evaluate $\int \frac{e^z}{z^2+ \pi^2}dz$ where C is |z|=4 ,using residue theorem
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The closed curve C is the circle with centre (0,0) & radius 4.

Poles are given by $z^2+π^2 = 0$ i.e. z = πi , - πi both lie inside C

Residue at the pole z= - πi is given as

$R_1 = lim_{z→ (-πi)}⁡ \,\ (z+πi ) \frac{e^z}{(z+πi)(z-πi)} = lim_{z→ (-πi)} \,\ ⁡\frac{e^z}{(z-πi)}$

=$\frac{e^{-πi}}{(-πi-πi)} = - \frac{e^{-πi}}{2πi} = \frac{1}{2πi}$

Also Residue at the pole z= - πi is given as

$R_2 = lim_{z→ πi} \,\ ⁡(z-πi ) \frac{e^z}{(z+πi)(z-πi)} = lim_{z→ πi} \,\ \frac{e^z}{(z+πi)}$

= $\frac{e^{πi}}{(πi+πi)} = \frac{e^{πi}}{2πi} = \frac{-1}{2πi}$

By residue theorem value of the integral

$∫ \frac{e^z}{z^2+π^2} = 2πi[ R_1 + R_2 ] = 2πi[\frac{1}{2πi}-\frac{1}{2πi}] = 0$