Factoring polynomial with complex coefficients 
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

What you did is correct.
You can check by substitution:  If $z=3-2i$ then
$$(z+2i)^2-9= ((3-2i)+2i)^2-9 = 3^2-9=0.$$
If $z=-3-2i$ then
$$
(z+2i)^2 - 9 = ((-3-2i) + 2i)^2 - 9 = (-3)^2 - 9 = 0.
$$
