Question

Evaluate the $\oint$(12z-7)/(z-1)2(2z+3) dz where c is the circle |z+i|= Sqrt3

Answer 1

Singular points of the integrand:

${z_1} = 1$ - second order pole,$\,\,{z_2} = - \frac{3}{2}$ - simple pole.

Only a point ${z_1} = 1$ falls inside the circle $|z + i| = \sqrt 3$

Let's find the residue at this point

$\mathop {res}\limits_{z = 1} f(z) = \mathop {\lim }\limits_{z \to 1} {\left( {\frac{{12z - 7}}{{{{(z - 1)}^2}\left( {2z + 3} \right)}}{{\left( {z - 1} \right)}^2}} \right)^\prime } = \mathop {\lim }\limits_{z \to 1} {\left( {\frac{{12z - 7}}{{2z + 3}}} \right)^\prime } = \mathop {\lim }\limits_{z \to 1} \frac{{12(2z + 3) - (12z - 7) \cdot 2}}{{{{\left( {2z + 3} \right)}^2}}} = \frac{{12 \cdot 5 - 2\left( {12 - 7} \right)}}{{{5^2}}} = 2$

Then

$\int\limits_{|z + i| = \sqrt 3 } {\frac{{12z - 7}}{{{{(z - 1)}^2}\left( {2z + 3} \right)}}dz} = 2\pi i\sum {\mathop {res}\limits_{z = {z_k}} } f(z) = 2\pi i \cdot 2 = 4\pi i$

Answer:$4\pi i$