Series expansion for innocent looking function

Question

$$f(z) = \frac{1}{z^2}$$ is given, where $f(z)$ is complex valued function.

How can one find series expansion at $ z=i$ with using geometric series approach? It seems simple but first tries gives no result, i.e adding subtracting $i$.

Answer 1

There are two ways to compute the power series. In each way you need to compute the series for $g(z) = 1/z$ using the geometric approach.

$$g(z) = \frac{1}{z} = -\frac{1}{i - (z-i)} = i \frac{1}{1 - (\frac{z-i}{i})} = \sum_{k=0}^{\infty} (-i)^{k-1} (z-i)^k$$

for $|z - i| < 1$. Now you have 2 options (I personally prefer option number 2):

1. Use the Cauchy product, i.e. use $$\left( \sum_{n=0}^{\infty} a_n \right) \cdot \left( \sum_{n=0}^{\infty} b_n \right) = \sum_{n=0}^{\infty} \sum_{k=0}^n a_k b_{n-k}$$

2. Use a theorem about the derivatives of power series. The theorem I'm thinking of tells you that taking the derivative of a power series which converges absolutely can be taken order-by-order. Doing this, you get $$\frac{1}{z^2} = - \frac{d}{dz} g(z)$$

Both approaches yield the result

$$\frac{1}{z^2} = \sum_{n=0}^{\infty} (-1)(-i)^n (n+1) (z-i)^n$$