# cauchy's integral theorem

• Apr 10th 2010, 03:39 AM
Tekken
cauchy's integral theorem
Im attempting to show that the following is not true by cauchy's integral theorem.

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

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

I know i need to do something with the $\overline{z}$ but cant figure out what
• Apr 10th 2010, 03:45 AM
Laurent
Quote:

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

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

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

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

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

Anyway, the trick is to notice that, if $z\in C$, then $\overline{z}=\frac{1}{z}$, so you can apply Cauchy theorem and there is a pole inside the contour.
• Apr 10th 2010, 03:54 AM
Tekken
ya sorry about that, it should have been $z^{2} - 5z + 6$

Thanks for the help (Clapping)