Just verify!

$\begin{aligned} z & = \frac 1i + \frac 2{i^2} + \frac 3{i^3} + \frac 5{i^4} & \small \color{#3D99F6}{\text{Multiply throughout by }1=i^4} \\ & = i^3 + 2i^2 + 3i + 5 \\ & = \color{#3D99F6}{(i^2)}i + 2\color{#3D99F6}{(i^2)} + 3i + 5 & \small \color{#3D99F6}{\text{Note that }i^2=-1} \\ & = \color{#3D99F6}{-i-2} + 3i + 5 \\ & = \boxed{2i+3} \end{aligned}$

$i^{2}=-1,i^{3}=-i,i^{4}=1$ where $\sqrt{-1}=i$ Then,

$\frac{1}{i}+\frac{2}{i^{2}}+\frac{3}{i^{3}}+\frac{5}{i^{4}}=\frac{1}{i}+\frac{2}{-1}+\frac{3}{-i}+\frac{5}{1}$

$=\frac{1}{i}-\frac{3}{i}-2+5$

$=\frac{-2}{i}+3=\boxed{2i+3}$

$a + ib = \dfrac{1}{i} + \dfrac{2}{i^2} + \dfrac{3}{i^3} + \dfrac{5}{i^4} \\ = \dfrac{1}{i} \times \dfrac{i}{i} + \dfrac{2}{-1} + \dfrac{3}{-i} + \dfrac{5}{1} \\ = -i -2 + 3i + 5 \\ = 3 + 2i$