Solve equation with complex numbers using a helper equation

Question

For the last two hours I've been trying to solve this complex equation using a helper equation. But I can't work it out.

$z^2 = 5-12$

$\text{Let} \space z = x + yi$

$(x+yi)^2 = 5-12i$

$x^2-y^2 + 2ixy = 5-12i$

This means that we get the following system of equations:

$\begin{cases} x^2-y^2 = 5 \\ 2xy = 12 \end{cases}$

To solve our system of equations I'm supposed to use something called a helper equation that uses the absolute value of the complex numbers:

$|z^2| = \sqrt{5^2-12^2} = 13$

$\text{If} \space z = x+yi \space \text{then} \space |z|^2 = 13 \Leftrightarrow x^2+y^2 = 13 $

At this point I can't really follow my notes from the lecture, I think they are not complete. How would I continue from here? If I'm not mistaken my lecturer set $w = z + something$, and after knowing what $w$ was he could solve what $z$ should be.

Answer 1

HINT:

$$(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2=5^2+12^2=13^2\implies x^2+y^2=13$$

We already have $x^2-y^2=5$

As $xy=6>0;x,y$ must be of same sign

Answer 2

You can solve the system $$\left\{\begin{array}{c} x^2+y^2=13\\x^2-y^2=5\end{array}\right.$$ for $x^2$ and $y^2$. This just gives you a very few possibilities for $x$ and $y$, which must also satisfy $xy=6$.