# cauchy's integral theorem

• Apr 10th 2010, 02:39 AM
Tekken
cauchy's integral theorem
Im attempting to show that the following is not true by cauchy's integral theorem.

$\displaystyle \int_{C}\frac{\overline{z}}{z^{2} - 5z - 6},dz = 0$ where $\displaystyle C$ is the unit circle (travelled anti-clockwise).

Resolving $\displaystyle z^{2} - 5z - 6 = 0$ i got z = 2 and z = 3 but these singularities lie outside the unit circle...

I know i need to do something with the $\displaystyle \overline{z}$ but cant figure out what
• Apr 10th 2010, 02:45 AM
Laurent
Quote:

Originally Posted by Tekken
Im attempting to show that the following is not true by cauchy's integral theorem.

$\displaystyle \int_{C}\frac{\overline{z}}{z^{2} - 5z - 6},dz = 0$ where $\displaystyle C$ is the unit circle (travelled anti-clockwise).

Resolving $\displaystyle z^{2} - 5z - 6 = 0$ i got z = 2 and z = 3 but these singularities lie outside the unit circle...

I know i need to do something with the $\displaystyle \overline{z}$ but cant figure out what

Either it is $\displaystyle z^2-5z+6$ or you did a mistake in computing the roots.

Anyway, the trick is to notice that, if $\displaystyle z\in C$, then $\displaystyle \overline{z}=\frac{1}{z}$, so you can apply Cauchy theorem and there is a pole inside the contour.
• Apr 10th 2010, 02:54 AM
Tekken
ya sorry about that, it should have been $\displaystyle z^{2} - 5z + 6$

Thanks for the help (Clapping)