Algebra Precalculus – How to Solve z^2-(1-3i)z-2i-2=0

algebra-precalculuscomplex numbersquadratics

I've tried two ways, but get stuck.

I've tried to simplify, but didn't know what to do next, and I've tried to solve it like a Quadratic equation but got stuck too.

One way got me this:

$$\frac{z}{2} \times (-1+3i+z)-1-i=0$$ – don't know what next.

option b, got me mess, while trying to make the equation into a Quadratic.

Best Answer

Using the quadratic formula for a general quadratic $ax^2+bx+c=0$ which is $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=0$$ for $z^2-(1-3i)z-2i-2=0$ $$z=\frac{1-3i\pm \sqrt{-8-6i+8i+8}}{2}$$ $$\implies z=\frac{1-3i\pm \sqrt{2i}}{2}$$