Complex numbers equation: $z^4 = -16$ Probably dumb question, but I would like to ask it anyway.
I was to solve this equation:
$z^4 = -16$
At first glance, the way to solve it would be (as any other equation):
$z^2 = \sqrt{-1} * 4 \lor z^2 = -\sqrt{-1} * 4$
$z^2 = 4i \lor z^2 = -4i$
$z = 2\sqrt{i} \lor z = -2\sqrt{i} \lor z = 2i\sqrt{i} \lor z = -2i\sqrt{i}$
However, in suggested answers, there are $z_1, z_2, z_3$ and $z_4$ found using trigonometric formulas. Is there something particulary wrong in my solution and the only proper one uses trigonometric form of complex number?
 A: Short way: polar coordinates.
$$
z^4=-16=(-1)2^4=2^4e^{i\pi}=2^4e^{i\pi+2k\pi i}=2^4e^{i\pi(2k+1)}\;,\;\;k\in\mathbb Z\;\;
$$
from which you get
$$
z=2e^{i\frac\pi 4(2k+1)}\;,\;\;k\in\mathbb Z
$$
which assumes only $4$ different values as $k\in\mathbb Z$, so we can write
$$
z_k=2e^{i\frac\pi 4(2k+1)}\;,\;\;k=0,1,2,3
$$
that are the four $4$-th roots of $-16$.
A: Say we want the two square roots of $4i$.  The angle between the positive real axis and the line on which $4i$ lies is $90^\circ$ or $-270^\circ$.  For the square root, you want half that angle, since angles are added when complex numbers are multiplied.  It's enough to find one of the two square roots, since after that we can just write "$\pm$" in front of it.  Half of $90^\circ$ is $45^\circ$, so we want $a + ai$, i.e. the real and imaginary parts are equal to each other when the angle is $45^\circ$.  The absolute value $|a+ai$ is $\sqrt{a^2+a^2}= a\sqrt{2}$ ($a$ is positive because of what we said earlier).  We want the absolute value to be $2$ so that the absolute value of the square will be $4$.  So we need $a\sqrt{2} = 2$.  Therefore $a=\sqrt{2}$.
The two square roots of $4i$ are therefore $\pm(\sqrt{2} + i\sqrt{2})$.
Use a similar method for $-4i$.
A: We have
$$z^4 + 16 = (z^2 + 4)^2 - 8z^2 = (z^2 - 2\sqrt{2} z + 4)(z^2 + 2\sqrt{2} z + 4).$$
Anotherway $$z^2 - 2\sqrt{2} z + 4 = (z-\sqrt{2})^2 + 2 = (z-\sqrt{2})^2 -(\sqrt{2} i)^2 =  (z-\sqrt{2} + \sqrt{2} i)(z-\sqrt{2} + \sqrt{2} i)).$$
The same for $z^2 + 2\sqrt{2} z + 4$
A: Therefore, if you continue on from @Joe's proposed solution, you should get 
Z1=$2cis(\pi/4)=\sqrt2+\sqrt2i$; 
Z2=$2cis(3\pi/4)=-\sqrt2+\sqrt2i$; 
Z3=$2cis(5\pi/4)=-\sqrt2-\sqrt2i$; 
Z4=$2cis(7\pi/4)=\sqrt2-\sqrt2i$.
If you check, it also makes sense, because complex numbers come in conjugate pairs. Hope it helps. 
