Question

Find the disk of convergence for each of the following complex power series.

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{\left(}\mathbf{i}\mathbf{z}\mathbf{\right)}}^{\mathbf{n}}}{{\mathbf{n}}^{\mathbf{2}}}$

Short answer

The required disk of convergence is .$\left|z\right|<1$

Step-by-step solution

Determine Disk of Convergence

For any power series$\sum a_n{z}^n$ where z is a complex numbers, then disk of convergence is given by:.$\mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathbf{\left|}\frac{\mathbf{z}\mathbf{\times }\mathbf{n}}{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\right|}\mathbf{=}\mathbf{\left|}\mathbf{z}\mathbf{\right|}$

Step 2:Find the disk of Convergence

The given power series is:,$\sum _{n=1}^{\infty }\frac{{\left(iz\right)}^n}{n^2}$where, .$a_n=\frac{{\left(iz\right)}^n}{n^2}$

Now, let us evaluate the ratio as:

$\begin{array}{c}\rho =\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{\frac{{\left(iz\right)}^{n+1}}{{(n+1)}^2}}{\frac{{\left(iz\right)}^n}{n^2}}\right|\\ =\underset{n\to \infty }{\lim }\left|iz{\left(\frac{n}{n+1}\right)}^2\right|\end{array}$

Now, for the series to be convergent, we have .$\rho <1$So,

$\begin{array}{c}\rho =\underset{n\to \infty }{\lim }\left|iz{\left(\frac{n}{n+1}\right)}^2\right|<1\\ \left|iz\right|<\underset{n\to \infty }{\lim }\left|{\left(\frac{n}{n+1}\right)}^2\right|\\ \left|z\right|<\underset{n\to \infty }{\lim }\left|{\left(\frac{n}{n+1}\right)}^2\right|\\ \left|z\right|<1\end{array}$

Hence, the required disk of convergence is .$\left|z\right|<1$