# How do you find a power series representation for  f(z)=z^2  and what is the radius of convergence?

Nov 20, 2015

$f \left(z\right)$ is effectively a power series already with a radius of convergence of $\infty$.

#### Explanation:

A power series (centered at $0$) is just a sum of the form $f \left(z\right) = {\sum}_{n = 0}^{\infty} {a}_{n} {z}^{n}$
so in this case, $f \left(z\right) = {z}^{2}$ is already a power series with
${a}_{n} = \left\{\begin{matrix}1 \mathmr{if} n = 2 \\ 0 \mathmr{if} n \ne 2\end{matrix}\right.$

In general, no manipulation is needed to find the power series of a polynomial function, as a power series is itself essentially a polynomial of infinite degree.

As for the radius of convergence, for any real value, the above power series has a single nonzero term which is equal to the square of that value, and thus does not diverge. This means the radius of convergence is infinite.