Sum of complex numbers

$\begin{aligned} S & = \frac 1z + \frac 2{z^2} + \frac 3{z^3} + \frac 4{z^4} + ... \\ & = \sum_{\color{#3D99F6}n=1}^\infty \frac n{z^n} \\ & = \sum_{\color{#D61F06}n=0}^\infty \frac n{z^n} \\ & = \sum_{\color{#3D99F6}n=1}^\infty \frac {n+1}{z^{n+1}} \\ & = \frac 1z \sum_{\color{#3D99F6} n=1}^\infty \frac n{z^n} + \frac 1z \sum_{n=1}^\infty \frac 1{z^n} \\ & = \frac 1z S + \frac 1z \sum_{n=1}^\infty \frac 1{z^n} \\ (z-1)S & = \sum_{n=1}^\infty \frac 1{z^n} = \frac 1{1-\frac 1z} = \frac z{z-1} \\ \implies S & = \frac z{(z-1)^2} = \frac {1-3i}{(-3i)^2} = \boxed{\dfrac {3i-1}9} \end{aligned}$

Let $\large x= \frac {1}{z}+\frac{2}{z^2}+\frac{3}{z^3}+\frac{4}{z^4}+...$ ;

Multiply $\large x \ by \ z$ to get $\large xz= 1+ \frac{2}{z}+\frac{3}{z^2}+\frac{4}{z^3}+...$

Subtract $\large xz -x = 1+ \frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+...$ , which is a geometric series with $\large a_1= 1 and\ r=\frac{1}{z}$ ,

thus, $\large xz-x= \frac{1}{1-\frac{1}{z}} = \large \frac{z}{(z-1)}$ , (the infinite sum of a geometric series = $\large\frac{a_1}{1-r}$ ),

solving for $\large x$ we get $\large x= \large \frac{z}{(z-1)^2}$ .

Plugging in the value of $\large z = 1-3i$ , we get $\large x= \frac{1-3i}{(1-3i-1)^2}=\boxed{ \frac {3i-1}{9}}$