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.