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$\oint \frac{dz}{z^3(z+4)}$ ,where c is the circle |z|=2.
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The path of integration is the circle with centre (0,0) & radius =2

The points at which the function is not analytic are z= 0 & 4 out of these z = 0 lies inside the circle & z =4 lies outside C .

By Cauchy’s Integral formula value of the integral is given by

$∮ \frac{dz}{z^3 (z+4)} = ∮ \frac{1/(z+4)}{z^3} \ dz$

= $\frac{2πi}{2!} [\frac{d^2}{dz^2} (z+4)^{-1} ]_{z=0}$

= $πi [\frac{d^2}{dz^2} (z+4)^{-1} ]_{z=0}$

= $- πi [\frac{d}{dz} (z+4)^{-2} ]_{z=0} = 2 πi [ (z+4)^{-3} ]_{z=0}$

= $2 \frac{πi}{64} = \frac{πi}{32}$