How to solve for all z: $z^2+i\sqrt{32}z+6i$

Question

this is my first question and I don't quite understand how do I confront this equation:

$z^2+i\sqrt{32}z-6i=0$

I tried using the quadratic formula but it doesn't seem to give me a correct answer, any help will be much obliged.

Thank you! :)

Answer 1

if z is a complex number let $z=a+bj$,

$$(a+bj)^2+\sqrt{32}(a+bj)j-6j=0$$ $$\therefore a^2-b^2+2abj+\sqrt{32}aj-\sqrt{32}b-6j=0$$ $$\therefore a^2-b^2-\sqrt{32}b=0, 2abj+\sqrt{32}aj-6j=0$$ These can then be solved simultaneously to find the roots

Answer 2

Hint: Let $z = a + ib$, separate the equation into the real and imaginary parts and equate both parts to $0$.

Answer 3

Note that $z^2+i\sqrt{32}z-6i=(z+i\sqrt{8})^2+(8-6i)$. The problem is essentially$$(z+i\sqrt{8})^2+(8-6i)=0$$