How to expand $\frac{1}{(1+z^2)}$ in powers of $(z-a)$?

Question

How to expand $\frac{1}{(1+z^2)}$ in powers of $(z-a)$?

I know for people who knows how to do this this is a stupid problem. But I am just a beginner. Differentiating $\frac{1}{(1+x^2)}$ seems not a sensable way. I also try to do this problem from the taylor series at $0$ but failed because I cannot see a clear pattern.

Thank you for your help!

Answer 1

From the binomial theorem,

$$(1+x)^{-1}=\sum_{n=0}^{\infty}(-1)^nx^n$$

Letting $x=z^2$,

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

This gives the series expansion around $z=0$. To expand around $z=a$, you can try this:

As Mayank Pandey pointed out,

$$\frac{1}{z^2 + 1}=\frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right)$$

For the first term,

$$\begin{array}{ccl} \frac{1}{z+i}&=&\frac{1}{(a+i)+(z-a)}\\ &=&\frac{1}{a+i}\frac{1}{1+\frac{z-a}{a+i}}\\ &=&\frac{1}{a+i}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z-a}{a+i}\right)^n \end{array}$$

Do the same to the second term and combine like terms in the series.

Answer 2

$$\frac{1}{z^2 + 1} = \frac{1/2i}{z - i} - \frac{1/2i}{z + i}$$ In general, you can do this to any poynomial.