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Using Cauchys integral formula ,evaluate $\oint \frac{12z-7}{(z-1)^2(2z+3)}dz$, where c is the circle $|z+i| = \sqrt{3}$
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The path of integration is the circle with centre (0,-1) & radius = $\sqrt{2}$

The points at which the function is not analytic are z= 1 & $\frac{-3}{2}$ both the points lie inside the circle C . Therefore by partial fraction method first we do the partial fraction of:

$\frac{1}{(z-1)^2 (2z+3)} = \frac{(-2/25)}{(z-1)} + \frac{(1/5)}{(z-1)^2} + \frac{(4/25)}{(2z+3)}$

By Cauchy’s Integral formula value of the integral

$∮ \frac{(12z-7)}{(z-1)^2 (2z+3)} \ dz$

= $\frac{-2}{25} ∮\frac{(12z-7)}{(z-1)} \ dz+ \frac{1}{5} ∮\frac{(12z-7)}{(z-1)^2} \ dz +\frac{4}{25} ∮\frac{(12z-7)}{(2z+3)} \ dz$

= $\frac{-2}{25}.2πi [12 z-7]_{z=1} + \frac{1}{5}.2πi \frac{d}{dz} [12 z-7]_{z=1} + \frac{4}{25}.2πi [12 z-7]_{z=(-3/2)}$

= $2πi [- \frac{10}{25}+\frac{12}{5}-4] = - 4πi$