Question

Given that $z$ is a non zero complex number, $i{z}^2=1+\frac{2}{z}+\frac{3}{z^2}+\frac{4}{z^3}+\frac{5}{z^4}+\dots \dots .+\infty$ and $z=n\pm \sqrt{-i}$ find the value of $100n$.

Answer 1

Answer: 100

Let $S=1+\frac{2}{z}+\frac{3}{z^2}+\frac{4}{z^3}+\dots \dots$

$S=\frac{z^2}{(z-1{)}^2}$
$\therefore i{z}^2=\frac{z^2}{(z-1{)}^2}(z\ne 0)$
$i=\frac{1}{(z-1{)}^2}\Rightarrow (z-1{)}^2=\frac{1}{i}=-i\Rightarrow z-1=\pm \sqrt{-i}$
$z=1\pm \sqrt{-i}$
$\text{ given }z=n\pm \sqrt{-i}$
$\Rightarrow 100n=100$