Given the equation $z^2+4iz-13=0$, solve for $z$ without the quadratic formula.
In real numbers set, when I find this kind of equations I usually complete the perfect square trinomial.In this case:
$(z^2+4iz-4)-13+4=0$
$(z+2i)^2-9=0$
I chosen $-4$ because the number whose double is $4i$, is $2i$. And the square of $2i$ is $-4$.
$z+2i= \pm \sqrt{9}$
$z=3-2i \vee z=-3-2i$
Is this correct?Thanks