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Using Cauchy's integral formula, evaluate (12z7)(z1)2(2z+3) where C:|z+1|=3
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Let I=(12z7)(z1)2(2z+3)

Circle |z+1|=3 has centre (0,1) and radius 3

Here, Z0=32 lies outside while Z0=1 lies inside the side

Z0=1 in Q pole of order 2

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I=(12z7)(2z+3)(z1)2dz

$Let \space \space f(z)=\dfrac {12z-7}{2z+3}\space and \space \space Z_0=1\\ \therefore f^1(z)=\dfrac {(2z+3)12-(12\times 1-7)2}{(2z-3)^2}\\ \therefore f^1(z_0)=f^1(1)=\dfrac {2(2(1)+3)12-(12\times …

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