Let f be an antiholomorphic function in C. $z_0 \in C - C(0,1). $
Show that
$\frac{1}{2 \pi i}\oint \frac {f(z)}{z-z_0} = \begin{cases}f(0) &\text{for } |z_0| < 1\\f(0) - f(\frac{1}{z_0}) &\text{for } |z_0|>1 \end{cases}$
Does anyone have any idea? I'm still thinking about it but don't know how to do it.
$f(z)$ anti-holomorphic over $\mathbb{C}$ if and only if $f(z) = g(\bar{z})$ for some holomorphic function $g(z)$. Notice on the unit circle $C(0,1)$, $\bar{z} = \frac{1}{z}$, we have
$$ \frac{1}{2\pi i}\oint_{C(0,1)} \frac{f(z)}{z-z_0} dz = \frac{1}{2\pi i}\oint_{C(0,1)} \frac{g(\bar{z})}{z-z_0} dz = \frac{1}{2\pi i}\oint_{C(0,1)} \frac{g\left(\frac{1}{z}\right)z}{z - z_0}\frac{dz}{z}$$ Notice under change of variable $\displaystyle\;w = \frac{1}{z}$,
$\displaystyle\;\frac{dz}{z} = -\frac{dw}{w}$.
$C(0,1)$ get mapped to $C(0,1)$ but the orientation of the contour get flipped.
The negative sign get cancelled out and we have
$$ \begin{align} \frac{1}{2\pi i}\oint_{C(0,1)} \frac{f(z)}{z-z_0} dz &= \frac{1}{2\pi i}\oint_{C(0,1)} \frac{g(w)}{1-z_0 w}\frac{dw}{w}\\ &= \frac{1}{2\pi i}\oint_{C(0,1)} g(w)\left(\frac{1}{w} - \frac{1}{w - \frac{1}{z_0}}\right) dw\\ &= \begin{cases} g(0) - g(\frac{1}{z_0}), & |\frac{1}{z_0}| < 1\\ g(0), & |\frac{1}{z_0}| > 1 \end{cases}\\ &= \begin{cases} f(0), & |z_0| < 1\\ f(0) - f(\frac{1}{\bar{z}_0}), & |z_0| > 1 \end{cases} \end{align} $$ Please note that for the case $|z_0| > 1$ in last expression, it is $\displaystyle\;f\left(\frac{1}{\bar{z}_0}\right)$ instead of $\displaystyle\;f\left(\frac{1}{z_0}\right)$.