(z*) * z + 4iz + 4i = 0, complex equation

Question

$(z*) * z + 4iz + 4i = 0$

How to solve an equation like this? Substituting z with a + bi gives:

$z^2 + 4i(a+bi)+4i = 0$

$z^2+ 4ai - 4b + 4i = 0$

What to do next?

Answer 1

solving the equation $$z^2+4iz+4i=0$$ we get $$z_1=-2i+2\sqrt{-1-i}$$ or $$z_2=-2i-2\sqrt{-1-i}$$ we get this using the formula for a quadratic equation of the form $$x^2+px+q=0$$ which has the Solutions $$x_1=-\frac{p}{2}+\sqrt{\frac{p^2}{4}-q}$$ or $$x_2=-\frac{p}{2}-\sqrt{\frac{p^2}{4}-q}$$