<s>[INST] complex number equation I have this equation:
$$z^2=-i$$
All I can figure out from my knowledge is that 
$$z^2=r^2\cos(2x)+i\sin(\cos(2x))$$ in this case is:
$$\cos(3\pi/4)+i\sin(3\pi/4))$$
and that will be $$e^{3\pi/4}$$
because $-i =0-i$ so that $\cos2x=0$ and $\sin2x=-1$.
Am I on the right track? I feel like I'm missing something.
The answer should be $\pm(-1+i)/\sqrt{2}$.
where does the $\sqrt{2}$ come from?
 [/INST] Different ways to represent complex numbers:
$z = x + iy$
$z = \rho (\cos\theta + i \sin \theta)$
$z = \rho e^i\theta$
$z^2 = (x^2-y^2) + i(2xy)$
$z^2 = \rho^2 (\cos2\theta + i \sin 2\theta)$
$z^2 = \rho^2 e^{2i\theta}$
This is important background, and if you are not comfortable bouncing between these, then keep reviewing until it clicks.
lets use this one:
$z^2 = \rho^2 (\cos2\theta + i \sin 2\theta) = -i\\
z^2 = (\cos \frac {3\pi}{2} + i \sin \frac {3\pi}{2}) = -i\\
2\theta = \frac {3\pi}{2}\\
\theta = \frac {3\pi}{4}\\
z = (\cos \frac {3\pi}{4} + i \sin \frac {3\pi}{4})\\
z = -\frac {\sqrt 2}{2} + i \frac {\sqrt 2}{2})\\
z = \frac{\sqrt 2}{2} (-1 + i)$
$z^2 = -i$ has two solutions.  You can choose to think if it as $\pm$ as was so handy with real numbers.  But with complex numbers, I think of it as rotating clockwise and counter clockwise.
So I think of the second solution as:
$z = (\cos \frac {7\pi}{4} + i \sin \frac {7\pi}{4})$ that is half-way between $\frac {3\pi}{2}$ and $2\pi$
$z = \frac{\sqrt 2}{2} (1 - i)$
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