cauchy's integral theorem

**Tekken** · Apr 10th 2010, 02:39 AM

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

**Laurent** · Apr 10th 2010, 02:45 AM

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.

**Tekken** · Apr 10th 2010, 02:54 AM

ya sorry about that, it should have been $z^{2} - 5z + 6$

Thanks for the help

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