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 8.8 years ago by

teamques10 ★ 69k

• modified 8.8 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

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$\therefore I=\int\dfrac {\frac {(12z-7)}{(2z+3)}}{(z-1)^2}dz$

$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 …

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