how to factorise an equation involving complex nos (cubic eqn)

z^​3​i+z²​-z+i=0

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\mathrm{iz}}^3+{\mathrm{z}}^2-\mathrm{z}+\mathrm{i}=0 \\ \Rightarrow {\mathrm{iz}}^2\left(\mathrm{z}+\frac{1}{\mathrm{i}}\right)-1\left(\mathrm{z}-\mathrm{i}\right)=0 \\ \mathrm{As} \frac{1}{\mathrm{i}}=-\mathrm{i}, \mathrm{hence} \mathrm{we} \mathrm{get} \\ \Rightarrow {\mathrm{iz}}^2\left(\mathrm{z}-\mathrm{i}\right)-1\left(\mathrm{z}-\mathrm{i}\right)=0 \\ \Rightarrow \left(\mathrm{z}-\mathrm{i}\right)\left({\mathrm{iz}}^2-1\right)=0 \\ \Rightarrow \mathrm{z}-\mathrm{i}=0 \mathrm{or} {\mathrm{iz}}^2-1=0 \\ \Rightarrow \boxed{\mathrm{z}=\mathrm{i}} \mathrm{or} {\mathrm{z}}^2=\frac{1}{\mathrm{i}} \\ \mathbf{\Rightarrow }\boxed{\mathbf{z}\mathbf{=}\mathbf{i}}\mathbf{ }\mathbf{or}\boxed{\mathbf{ }{\mathbf{z}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{i}} \end{gathered}$$

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