Solve$$\int_{|z|=5} \frac{z^2}{(z-3i)(z-3i)}.$$
So I am currently working on the unit about the Cauchy Integral Formula, where$$\int\frac{f(z)}{z-z_0}=2\pi i f(z_0).$$
The zero $z=3i$ lies within the circle $|z|=5$ so $z_0=3i$ and we can rewrite the integral so$$\int_{|z|=5} \frac{\frac{z^2}{z-3i}}{z-3i},$$where $f(z)=\frac{z^2}{z-3i}$. However, when I attempt to plug $z_0$ into $f(z)$, the denominator is zero which is definitely wrong.
Where am I going wrong?
There is this generalization of Cauchy's integral formula that you can apply here: under the same conditions as Cauchy's integral formula, you have, for each $n\in\Bbb Z_+$,$$\frac1{2\pi i}\oint_{|z-a|=r}\frac{f(z)}{(z-a)^{n+1}}\,\mathrm dz=\frac{f^{(n)}(a)}{n!}.$$So, take $n=1$ here. Of course, with $n=0$ this is just the standard formula.