$z^2=-3-3i$ solve for $z$ Use De Moivre's theorem to solve the equation $z^2=-3-3i$. (Give your answers in polar form)
Can you please explain why there are two answers? I cannot seem to understand why.
By the way, the answers are
$$z=18^{1/4}\mathrm{cis}(-3\pi/8)$$
$$z=18^{1/4}\mathrm{cis}(5\pi/8)$$
Thanks for the help.
 A: Notice that $-3 - 3i$ is equivalent to $3\sqrt{2}e^{-\frac{3\pi}{4}i}$.  To see how, write $-3-3i$ as $3(-1 - i)$.  Look at
$$-1 - i = re^{i\theta}$$
where $\theta$ is the angle and $r$ is the radius.  Then
$$\begin{aligned}
-1 &= r\cos(\theta)\\
-1 &= r\sin(\theta)
\end{aligned}$$
With some algebra, we have
$$\begin{aligned}
(r\cos(\theta))^2 + (r\sin(\theta)^2) &= (-1)^2 + (-1)^2\\
r^2(\cos^2(\theta) + \sin^2(\theta)) &= 2\\
r^2 &= 2\\
r &= \sqrt{2}\\
\Rightarrow -1 &= \sqrt{2}\cos(\theta)\\
\Rightarrow -1 &= \sqrt{2}\sin(\theta)\\
\Longrightarrow \theta &= -\dfrac{3\pi}{4}
\end{aligned}$$
So we have $3\sqrt{2}e^{-\frac{3\pi}{4}i}$, which by Euler's identity is equivalent to
$$3\sqrt{2}\left(\cos\left(-\dfrac{3\pi}{4}\right) + i\sin\left(-\dfrac{3\pi}{4}\right)\right)$$
which gives us
$$\begin{aligned}
z^2 &= 3\sqrt{2}\left(\cos\left(-\dfrac{3\pi}{4} \right) - i\sin\left(\dfrac{3\pi}{4}\right)\right)\\
z &= \left(3\sqrt{2}\left(\cos\left(-\dfrac{3\pi}{4} \right) - i\sin\left(\dfrac{3\pi}{4}\right)\right) \right)^{\frac{1}{2}}\\
\end{aligned}$$
with the following solutions found by De Moivre's Theorem:
$$\begin{aligned}
z_1 &= (3\sqrt{2})^{\frac{1}{2}}\left(\cos\left(\dfrac{-\frac{3\pi}{4}}{2} \right) + i\sin\left(-\dfrac{\frac{3\pi}{4}}{2} \right) \right)\\
&= (3\sqrt{2})^{\frac{1}{2}}\left(\cos\left(-\dfrac{3\pi}{8} \right) + i\sin\left( -\dfrac{3\pi}{8}\right)\right)\\
&= (3\sqrt{2})^{\frac{1}{2}}e^{-\frac{3\pi}{8}i}\\
&= (\sqrt{18})^{\frac{1}{2}}e^{-\frac{3\pi}{8}i}\\
&= 18^{\frac{1}{4}}e^{-\frac{3\pi}{8}i}\\
z_2 &= (3\sqrt{2})^{\frac{1}{2}}\left(\cos\left(\dfrac{-\frac{3\pi}{4} + 2\pi}{2} \right) + i\sin\left(-\dfrac{\frac{3\pi}{4} + 2\pi}{2} \right) \right)\\
&= 18^{\frac{1}{4}}e^{\frac{5\pi}{8}i}
\end{aligned}$$
By mathematics2x2life's comment, those solutions are equivalent to the following:
$$\begin{aligned}
z_1 &= 18^{\frac{1}{4}}\mathrm{cis}\left(-\dfrac{3\pi}{8} \right)\\
z_2 &= 18^{\frac{1}{4}}\mathrm{cis}\left(\dfrac{5\pi}{8} \right)
\end{aligned}$$
References


*

*Cis $\theta$: http://math.wikia.com/wiki/Cis_%CE%B8

*De Moivre's Theorem: http://www3.ul.ie/~mlc/support/Loughborough%20website/chap10/10_4.pdf
A: $$\pm\sqrt[4]{18}\left(\frac{\sqrt{2-\sqrt{2}}}2-\frac{\sqrt{2+\sqrt{2}}}{2}i\right)$$
