Find all complex numbers 
such that the series
![\[ \sum_{n=1}^{\infty} \frac{1}{\left(1+|z|^2\right)^n}\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7be6bb396f84d3b8a80efa4050c1d5f2_l3.png "Rendered by QuickLaTeX.com")
converges.
This is a geometric series
![\[ \sum_{n=1}^{\infty} \frac{1}{\left( 1 + |z|^2 \right)^n} = \sum_{n=1}^{\infty} \left( \frac{1}{1+|z|^2} \right)^n. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-62ed24558f548b5181cc9ca1bdecbfb8_l3.png "Rendered by QuickLaTeX.com")
Therefore it converge if and only if
![\[ \frac{1}{1+|z|^2} < 1 \quad \implies \quad |z|^2 > 0 \quad \implies \quad z \neq 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-6f486b81f23baafbdd45abab924345ec_l3.png "Rendered by QuickLaTeX.com")
Hence, the series converges for all 
except 
.
Find all complex numbers such that the series
converges.
This is a geometric series
Therefore it converge if and only if
Hence, the series converges for all except .