Applying Cauchy-Goursat theorem/ doing it the "hard way"

**arbolis** · July 20th 2010, 11:31 AM

I must apply Cauchy-Goursat theorem to show that $\int _C f(z)dz=0$ where $C:=\{ z\in \mathbb{C} ; |z|=1 \}$ and $f(z)=\frac{z^2}{z-3}$ .

So what I wrote was "f is analytic over $\mathbb{C}-\{ 3 \}$ " therefore according to C-G's theorem the line integral is worth 0 since the singularity of f isn't inside the enclosed (by C) domain. Yet I don't feel I've "showed" anything. Do you think it's enough?

So I even tried to directly show that $\int _{|z|=1} \frac{z^2}{z^3-3} dz=0$ . I parametrized the unit circle as $\gamma (\theta)=e^{i \theta}$ for $-\pi <\theta <\pi$ . I came up with that this integral is worth $i \int _{-\pi} ^{\pi} \frac{e ^{i 3 \theta}}{e^{i \theta}-3} d\theta$ . Then I've used Euler's formula and couldn't show anything. (I get a very long and ugly impossible thing). How would you show that the integral is worth 0 using only the definition of a line integral?

**Ackbeet** · July 20th 2010, 11:53 AM

All you have to do is show that f(z) is analytic inside your closed contour (which it is, because the only singularity is outside the contour). Then you simply quote the Cauchy-Goursat theorem, and you're done.

**chisigma** · July 20th 2010, 12:01 PM

Setting $z= e^{i \theta}$ the integral becomes...

$\displaystyle \int_{C} f(z)\ dz = i \ \int_{-\pi}^{\pi} \frac{e^{3 i \theta}}{e^{3 i \theta} -3}\ d \theta = - \frac{i}{3} \ \int_{-\pi}^{\pi} \frac{e^{3 i \theta}}{1 - \frac{e^{3 i \theta}}{3}}\ d \theta$ (1)

Now because $|\frac{e^{3 i \theta}}{3}| < 1$ is...

$\displaystyle \frac{1}{1 - \frac{e^{3 i \theta}}{3}} = \sum_{n=0}^{\infty} \frac{e^{3 i n \theta}}{3^{n}}$ (2)

... so that is...

$\displaystyle \int_{C} f(z)\ dz = -\frac{i}{3} \ \sum_{n=0}^{\infty} \int_{- \pi}^{\pi} \frac{e^{3 (n+1) i \theta}}{3^{n}}\ d \theta$ (3)

But any term of the series (3) is equal to 0 so that is...

$\displaystyle \int_{C} f(z)\ dz = 0$ (4)

Kind regards

$\chi$ $\sigma$

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