Solution:

Given, $\oint_c \frac{e^{2z}}{ (z+1)^4} dz$

$z=-1 \ is \ a \ pole \ of \ order \ 4 \ i.e.m=4$

$residue \ of \ f(z) \ at \ z=-1$

$\frac{1}{(m-1)!}\lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^{m} f(z)]$

$\frac{1}{(4-1)!}\lim_{z \to -1} \frac{d^{4-1}}{dz^{4-1}}[(z+1)^{4} \frac{e^{2z}}{ {(z+1)}^4} ]$

$\frac{1}{(3)!}\lim_{z \to -1} \frac{d^{3}}{dz^{3}}[e^{2z} ]$

$\frac{1}{6}[e^{2z} 8]$

=$\frac{8}{6}[e^{-2}]$

=$\frac{4}{3}[e^{-2}]$

$\oint_c \frac{e^{2z}}{ (z+1)^4} dz$ = $2\pi i f(z)$

$=2\pi i \frac {4}{3}e^{-2}$

$\oint_c \frac{e^{2z}}{ (z+1)^4} dz$=$\pi i \frac {8}{3}e^{-2}$