[SOLVED]rectangle around the pole

dwsmith

Well-known member
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
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.