Taylor Series of Complex Rational Function

Question

I attempted to derive a Taylor series for the complex function below (centered at $z_0$, with constant $a$). Does anyone know if this is correct, and would it be useful for finding Laurent coefficients or for anything in general?

$\frac{1}{(z+a)^{n+1}}= \sum\limits_{k=0}^{\infty} \frac{(-1)^k}{(z_0+a)^{k+1+n}} \frac{(k+n)!}{k! n!} (z-z_0)^k$

Answer 1

The result reported in the OP

$$\frac{1}{(z+a)^{n+1}}=\sum_{k=0}^\infty \frac{(-1)^k(z-z_0)^{k}}{(z_0+a)^{k+n+1}}\frac{(k+n)!}{n!\,k!}$$

is correct.

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HINTS TO ARRIVE AT THE RESULT:

$$\frac{1}{(z+a)^{n+1}}=\frac{(-1)^{n}}{n!}\frac{d^{n}}{dz^n}\left(\frac{1}{z+a}\right)$$

and for $|z-z_0|<|z_0+a|$

$$\frac{1}{z+a}=\frac1{z_0+a}\sum_{k=0}^\infty (-1)^k\left(\frac{z-z_0}{z_0+a}\right)^k$$