Easy way to solve $w^2 = -15 + 8i$ 
Solve $w^2=−15+8i$, where $w$ is complex.

Normally, I would convert this into polar coordinates, but the problem is that is too slow. 
What is another alternative? 
 A: Hint. Let $w=a+ib$ then you have to solve
$$a^2-b^2=-15,\quad ab=4.$$
A: Start out by letting $w = a + bi \implies w^2 = a^2-b^2 + 2abi = -15+8i \implies a^2-b^2 = -15, 2ab = 8 \implies a^2 - \dfrac{16}{a^2} = -15$. Put $x = a^2 \implies x - \dfrac{16}{x} + 15 = 0\implies x^2 + 15x - 16 = 0\implies (x-1)(x+16) = 0\implies x = 1\implies a = \pm 1, b = \pm 4$
A: A general method that will very often work is to write all variables as the sum of the real and imaginary parts, and then equate the real and imaginary parts of the equation.
In this case, we set $w = a + bi$, and we get
$$
(a + bi)^2 = -15 + 8i \\
$$
Then setting real and imaginary parts equal,
\begin{align*}
a^2 - b^2 &= -15 \tag{1} \\
2ab &= 8 \tag{2}
\end{align*}
From (2), $b = \frac{4}{a}$; plug into (1) and multiply by $a^2$ we get
$$
a^4 - 16 = -15a^2
$$
so
$$
(a^2 + 16)(a^2 - 1) = 0
$$
and $a, b \in \mathbb{R}$.
So $a = 1$ or $a = -1$. And that will make $b = 4$ or $b = -4$, respectively.
A: $$-15+8i=\\-16+8i+1=\\(4i)^2+8i+1=\\(4i+1)^2 \\ \to 
w^2=(4i+1)^2\\w=\pm(4i+1)$$
