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Evaluate $\int \frac{z^2}{(z-1)^2(z+1)}dz$ where C is |z| = 2, using residue theorem.
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The closed curve C is the circle with centre (0,0) & radius 2.

Poles are given by $(z-1)^2 (z+1)=0$ i.e.

z=1 is the pole of order 2 & z= -1 is the pole of order 1 ,both the poles lie inside the circle C.

Residue at the pole z= -1 is given as

$R_1 = lim_{z→ (-1)} \,\ (z+1). \frac{z^2}{(z-1)^2 (z+1)} = lim_{z→ (-1)} \,\ \frac{z^2}{(z-1)^2} = \frac{1}{4}$

Also Residue at the pole z=1 is given as

$R_2 = lim_{z→1} \,\ \frac{1}{1!} \frac{d}{dz} \ (z-1)^2 . \frac{z^2}{(z-1)^2 (z+1)} = lim_{z→1} \ \frac{d}{dz}⁡ (\frac{z^2}{z+1}) = \frac{3}{2}$

By residue theorem value of the integral

$∫ \frac{z^2}{(z-1)^2 (z+1)} = 2πi[ R_1 + R_2 ] = 2πi [\frac{1}{4}+\frac{3}{2}] = \frac{7πi}{2}$