Roots of a complex number I was studying for my maths quiz and I had some questions to ask about the answer 
The answer is provided here:

Here comes my question. 
1. How did [sqrt2/2 e^j3pi/4)]^-2/3 became [1/2 e^j3pi/4]^1/3?


*How did (2e^j-3pi/4)^1/3 become (2e^jpi/2)^1/3?


Thanks for the help!
 A: The question was to evaluate $(-\frac{1}{2}+j\frac{1}{2})^{-\frac{2}{3}}$  (where $j$ is the imaginary unit).
The solution key is correct up to a time but has an error where you see the whiteout as well as an error which somehow corrected his earlier error two lines later.  The locations of the errors are marked in red below but have been fixed in my post.  It was likely just a transcription error on the person's part who wrote up the solution.
$$\begin{array}{rll}
(-\frac{1}{2}+j\frac{1}{2})^{-\frac{2}{3}}&=[\frac{1}{2}(-1+j)]^{-\frac{2}{3}} & \text{factor out common} \frac{1}{2}\\
&=[\frac{1}{2}(\sqrt{2}e^{j\frac{3\pi}{4}})]^{-\frac{2}{3}}&\text{rewriting parenthesis as polar form}\\
&=(\frac{\sqrt{2}}{2}e^{j\frac{3\pi}{4}})^{-\frac{2}{3}}&\text{distribute the}~\frac{1}{2}\\
&=(\frac{1}{2}e^{j\frac{3\pi}{\color{red}{2}}})^{-\frac{1}{3}}&\text{distribute exponent of $2$ inside the parenthesis}\\
&=(2e^{-j\frac{3\pi}{2}})^{\frac{1}{3}}&\text{distribute exponent of $-1$ inside the parenthesis}\\
&=(2e^{j\color{red}{\frac{\pi}{2}}})^{\frac{1}{3}}&\text{by the fact}~-\frac{3\pi}{2}~\text{and}~\frac{\pi}{2}~ \text{are equivalent angles}\\
&=2^{\frac{1}{3}}e^{j\frac{\frac{\pi}{2}+2k\pi}{3}}~\text{for}~k=0,1,2&\text{By DeMoivre's}
\end{array}$$
