# Complex Integration

• Nov 9th 2009, 09:06 AM
Haven
Complex Integration
Let $\displaystyle f(z) = \bar{z}$ where $\displaystyle \bar{z}$ is the conjugate of z.

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

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

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

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

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

which brings us to:

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

How ever I cannot figure out what allows me to integrate here.
• Nov 9th 2009, 10:00 AM
chisigma
The residue theorem extablishes that...

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

... where...

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

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

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

... so that is...

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Nov 9th 2009, 10:33 AM
Haven
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.
• Nov 9th 2009, 02:35 PM
shawsend
First note if $\displaystyle z=0$ then $\displaystyle g(0)=\oint\frac{ds}{s^2}=0$ via $\displaystyle s=e^{it}$ and direct integration.
If $\displaystyle z\ne 0$ then $\displaystyle g(z)=\frac{1}{2\pi i z}\mathop\oint\limits_{|s|=1}\left(\frac{1}{s-z}-\frac{1}{s}\right)ds$.
$\displaystyle \mathop\oint\limits_{|s|=1}\frac{ds}{s-z}dz=\mathop\oint\limits_{s=z+e^{it}} \frac{ds}{s-z}=2\pi i$
$\displaystyle \mathop\oint\limits_{|s|=1}\frac{ds}{s}=2\pi i$
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 $\displaystyle |z|>1$, since then I would be deforming it across the singular point at $\displaystyle z$. So then $\displaystyle g(z)=0$ in the unit disc and certainly the zero function is analytic.