I'm having a bit of a problem on how to compute forward the values for z from this equation:
$$ z^3+z^2+iz^2-z+1+3i=0 $$
In the question there's given that one solution or root is z=i, by solving by polynomial division I get the equation $$ \begin{align} &z^2+z+2iz+i-3=0 \\ \iff & z^2+(1+2i)z+i-3=0\\ \iff &\left(z+\frac{1+2i}{2}\right)^2=z^2+(1+2i)z+\left(\frac{1+2i}{2}\right)^2. \end{align} $$ I substitute this expression with $w$. $$ \begin{align} &w^2-\left(\frac{1+2i}{2}\right)^2+i-3=0\\ \iff &\left(\frac{1+2i}{2}\right)^2=\frac{1}{4}+i+i^2=\frac{1}{4}+i-1\\ \iff & \left(-\frac{3}{4}+i\right)w^2-\left(\frac{3}{4}+i\right)+i-3\\ \iff &w^2+\frac{3}{4}-3=w^2=\frac{9}{4}\\ \end{align} $$ But it's from here where it struggles with me with the second degree equation for some reason. Could anyone explain how it should be done, I would really appreciate it, any form of trix or so.
Thanks in advance
You just have $\;w=\pm\dfrac{3i}2$, so $$z=-\frac{1+2i}2\pm\frac{3i}2=\cdots$$
Edit: As pointed by @WETutorialSchool, The final equation in $w$ is actually $$w^2=\frac 94,\quad\text{whence}\quad z=-\frac{1+2i}2\pm\frac{3}2=-(2+i),1-i.$$