Show that if $f$ is analytic on $B(z_0,r)\setminus\{z_0\}$, and $ \lim_{z\to z_0} f(z)(z-z_0)=A$ , then, $\oint_{|z-z_0|=r} f(z) dz=2\pi iA$

Question

Show that if $f$ is analytic on $B(z_0,r)\setminus\{z_0\}$ , and $$ \lim_{z\to z_0} f(z)(z-z_0)=A$$ then $$\oint_{|z-z_0|=r} f(z) dz=2\pi iA$$ for any $r$ : $0<r<R$

I am not sure how to start, first of all, it looks like the solution requires Cauchy's Formula, but since we don't know that $f$ is analytic in $z_0$, doesn't that make the domain not simply connected, and thus doesn't allow the use of the theorem?

Answer 1

Let$$g(z)=\begin{cases}(z-z_0)f(z)&\text{ if }z\ne z_0\\A&\text{ if }z=z_0.\end{cases}$$Then $g$ is continuous at $z_0$ and therefore, by Riemann's extension theorem, it is holomorphic. So, Cauchy's integral formula telles us that$$\frac1{2\pi i}\oint_{|z-z_0|=r}\frac{g(z)}{z-z_0}\,\mathrm dz=g(z_0)=A.$$So,$$\frac1{2\pi i}\oint_{|z-z_0|=r}f(z)\,\mathrm dz=A.$$