 Solution

Choose a small positively oriented circle $C_{\epsilon }$ about $z_0$ inside the given region. Cauchy's theorem on the region between the two contours givesContinuity makes the last integrand converge uniformly to $f(z_0)$ as $\epsilon \downarrow 0$. Its limit is $2\pi if(z_0)$, proving the Cauchy integral formula.

The same formula holds with an interior point $w$ near $z_0$. Subtract its value at $z_0$ and divide by $w-z_0$:The contour is a positive distance from $z_0$, so the integrands converge uniformly as $w\to z_0$. Passing to the limit gives the differentiated formulaOn the unit circle, $\bar{z}=1/z$ and $\mathrm{Im}z=(z-\bar{z})/(2i)$, proving the requested identity. Hence the requested integral isThe first integral is $2\pi i/3$. For the second useThe differentiated formula applied to the constant function gives $\int z^{-2}dz=0$; the remaining two terms integrate to $-6\pi i+6\pi i=0$. Therefore