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Using Cauchy's integral formula, evaluate $\int\dfrac {(12z-7)}{(z-1)^2(2z+3)}$ where $C:|z+1|=\sqrt3$
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written 7.9 years ago by | • modified 7.9 years ago |
Let $I=\int\dfrac {(12z-7)}{(z-1)^2(2z+3)}$
Circle $|z+1|=\sqrt3$ has centre $(0,-1)$ and radius $\sqrt3$
Here, $Z_0=\dfrac {-3}2$ lies outside while $ Z_0=1$ lies inside the side
$Z_0=1$ in Q pole of order 2
$$\therefore I=\int\dfrac {\frac {(12z-7)}{(2z+3)}}{(z-1)^2}dz$$
$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 …