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Using Cauchy's integral formula, evaluate ∫(12z−7)(z−1)2(2z+3) where C:|z+1|=√3
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written 8.7 years ago by | • modified 8.7 years ago |
Let I=∫(12z−7)(z−1)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
∴I=∫(12z−7)(2z+3)(z−1)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 …