
Complex Integration
Let $\displaystyle f(z) = \bar{z} $ where $\displaystyle \bar{z} $ is the conjugate of z.
Let $\displaystyle g(z) = \frac{1}{2\pi\\i}\int_{C}\frac{f(s)}{sz}\\ds$
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
$\displaystyle g(z) = \frac{1}{2\pi\\i}\int_{C}\frac{s^{2}}{s(sz)}\\ds$
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:
$\displaystyle g(z) = \frac{1}{2\pi\\i}\int_{C}\frac{1}{s(sz)}\\ds$
How ever I cannot figure out what allows me to integrate here.

The residue theorem extablishes that...
$\displaystyle \int_{C} f(s)\cdot ds= 2 \pi i \sum_{n} R_{n}$ (1)
... where...
$\displaystyle R_{n} = \lim_{s \rightarrow s_{n}} (ss_{n})\cdot f(s)_{s=s_{n}}$ (2)
... being $\displaystyle s_{n}$ is every pole of $\displaystyle f(*)$ inside the path C. In your case is...
$\displaystyle f(s)= \frac{1}{2 \pi i}\cdot \frac{1}{s\cdot (sz)}$ (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
$\displaystyle \chi$ $\displaystyle \sigma$

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.

How about thiis:
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 $\displaystyle g(z)=\frac{1}{2\pi i z}\mathop\oint\limits_{s=1}\left(\frac{1}{sz}\frac{1}{s}\right)ds$.
Now, you can directly integrate those. For the first, we can deform the path:
$\displaystyle \mathop\oint\limits_{s=1}\frac{ds}{sz}dz=\mathop\oint\limits_{s=z+e^{it}} \frac{ds}{sz}=2\pi i$
and:
$\displaystyle \mathop\oint\limits_{s=1}\frac{ds}{s}=2\pi i$
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.