Complex Integration

**Haven** · November 9th 2009, 09:06 AM

Let $f(z) = \bar{z}$ where $\bar{z}$ is the conjugate of z.

Let $g(z) = \frac{1}{2\pi\\i}\int_{C}\frac{f(s)}{s-z}\\ds$

Where C is the boundary of the unit circle. Show $g(z)$ is analytic in the interior of C by evaluating the integral.

By multiplying the top and bottom of the integrand by s, I get

$g(z) = \frac{1}{2\pi\\i}\int_{C}\frac{|s|^{2}}{s(s-z)}\\ds$

and since we are integrating on C and $C = \{ z = e^{i\theta} : 0\leq\theta\leq\\2\pi \}$ , then $|s|^{2} = 1$

which brings us to:

$g(z) = \frac{1}{2\pi\\i}\int_{C}\frac{1}{s(s-z)}\\ds$

How ever I cannot figure out what allows me to integrate here.

**chisigma** · November 9th 2009, 10:00 AM

The residue theorem extablishes that...

$\int_{C} f(s)\cdot ds= 2 \pi i \sum_{n} R_{n}$ (1)

... where...

$R_{n} = \lim_{s \rightarrow s_{n}} (s-s_{n})\cdot f(s)_{s=s_{n}}$ (2)

... being $s_{n}$ is every pole of $f(*)$ inside the path C. In your case is...

$f(s)= \frac{1}{2 \pi i}\cdot \frac{1}{s\cdot (s-z)}$ (3)

... so that is...

$g(z)= \frac{1}{z} - \frac{1}{z} =0$ , $|z|<1$

$g(z)= -\frac{1}{z}$ , $|z|>1$ (4)

If $|z|=1$ there is a pole is on the path C and the residue theorem can't be applied...

Kind regards

$\chi$ $\sigma$

**Haven** · November 9th 2009, 10:33 AM

As nice of an argument that is, we're only on integration in our complex variables course. So we're expected to give an argument that relies completely on integration methods. I love the residue theorem, but sadly I cannot use it.

**shawsend** · November 9th 2009, 02:35 PM

How about thiis:

First note if $z=0$ then $g(0)=\oint\frac{ds}{s^2}=0$ via $s=e^{it}$ and direct integration.
If $z\ne 0$ then $g(z)=\frac{1}{2\pi i z}\mathop\oint\limits_{|s|=1}\left(\frac{1}{s-z}-\frac{1}{s}\right)ds$ .
Now, you can directly integrate those. For the first, we can deform the path:

$\mathop\oint\limits_{|s|=1}\frac{ds}{s-z}dz=\mathop\oint\limits_{s=z+e^{it}} \frac{ds}{s-z}=2\pi i$
and:
$\mathop\oint\limits_{|s|=1}\frac{ds}{s}=2\pi i$
as well by the same substitution method. Note I could deform the path for the first one only because the deformation was over a region in which the function was analytic. I could not do this if $|z|>1$ , since then I would be deforming it across the singular point at $z$ . So then $g(z)=0$ in the unit disc and certainly the zero function is analytic.

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