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[SOLVED] rectangle around the pole

dwsmith

Well-known member
Feb 1, 2012
1,673
Find the integral
$$
\int_C\frac{dz}{z^2 - 3z + 5} = \int_C\frac{dz}{\left(z - \frac{3}{2}-i\frac{\sqrt{11}}{2}\right)\left(z-\frac{3}{2}+i\frac{\sqrt{11}}{2}\right)}
$$

Where the path is a rectangle oriented clockwise from (0,0) to (0,4) to (10,4) to (10,0) to (0,0).

So $z_1 = \frac{3}{2} + i \frac{ \sqrt{11} }{2}$ and $ z_2 = \frac{3}{2} - i \frac{ \sqrt{11} }{2}$

The $\int_C\frac{dz}{f(z)}=-2\pi i\text{Res}_{z_0}$

So the residue is
$$
\frac{1}{z_1-z_2} = \frac{1}{-i\sqrt{11}}
$$
Then
$$
\int_C\frac{dz}{z^2 - 3z + 5} = \frac{-2\pi i}{-i\sqrt{11}} =\frac{2\pi}{\sqrt{11}}
$$

Correct?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Find the integral
$$
\int_C\frac{dz}{z^2 - 3z + 5} = \int_C\frac{dz}{\left(z - \frac{3}{2}-i\frac{\sqrt{11}}{2}\right)\left(z-\frac{3}{2}+i\frac{\sqrt{11}}{2}\right)}
$$

Where the path is a rectangle oriented clockwise from (0,0) to (0,4) to (10,4) to (10,0) to (0,0).

So $z_1 = \frac{3}{2} + i \frac{ \sqrt{11} }{2}$ and $ z_2 = \frac{3}{2} - i \frac{ \sqrt{11} }{2}$

The $\int_C\frac{dz}{f(z)}=-2\pi i\text{Res}_{z_0}$

So the residue is
$$
\frac{1}{z_1-z_2} = \frac{1}{-i\sqrt{11}}
$$
Then
$$
\int_C\frac{dz}{z^2 - 3z + 5} = \frac{-2\pi i}{-i\sqrt{11}} =\frac{2\pi}{\sqrt{11}}
$$

Correct?
This is correct.