Find all complex numbers $z$ such that
$$z^2=12−16i,$$
and give your answer in the form $a+bi$.
We set
$$z= a+bi,$$
thus,
$$z^2 = (a^2 - b^2) + (2ab)i.$$
Equating both $z^2$ we have
$$ a^2 - b^2 = 12\text{ and }ab = -8.$$
I am told that I can find the answer by using the quadratic formula. However, I don't see a way how can I apply the quadratic formula with the given equation. quadratic formula works when we have
$$ax^2 + bx + c = 0$$
I don't know how do i apply this in the context of $z^2$.
Arguably the coolest (though not necessarily easiest) way to do with would be to use Euler's Identity. Set $12-16i=\cos(\theta)+i\sin(\theta)$, solve for $\theta$ using the fact that $\sqrt{(12+16i)(12-16i)}\cos(\theta)=12$. Then use the fact that $\cos(\theta/2)+i\sin(\theta/2)=\sqrt{\cos(\theta)+i\sin(\theta)}$ since $e^{i\theta}=\cos(\theta)+i\sin(\theta)$.
Since $ab=-8$, we have $b=-8/a$ so in the equation $$ a^2-b^2=12 $$ we can put $-8/a$ in place of $b$, getting $$ a^2-\frac{64}{a^2} = 12, $$ whence $$ a^4 - 64 = 12a^2 $$ or $$ c^2 - 64 = 12c. $$ That's an ordinary quadratic equation.
From $a^2-b^2=12$ and $ab=-8$ you can get $$a^2-\frac{64}{a^2}=12.$$ This can be converted to $$a^4-12a^2-64=0.$$ Use $t=a^2$ to get $$t^2-12t-64=0.$$ Now solve for $t$ (hence $a$ and $b$).
From $ab = -8$ we have $b = -\frac{8}{a}$. Substituting into the other equation $$ a^2 - \frac{8^2}{a^2} = 12 $$
Multiplying through by $a^2$ $$ a^4 - 12a^2 - 64 = 0 $$
You now have a quadratic equation in $a^2$
