<s>[INST] Finding square roots of $\sqrt 3 +3i$ I was reading an example, where it is calculating the square roots of $\sqrt 3 +3i$.
$w=\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)\\=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$
Let $z^2=w \Rightarrow r^2(\cos(2\theta)+i\sin(2\theta))=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$.
But how did they get from $\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$?
And can one just 'let $z^2=w$' as above?
Edit:
$w=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})=z^2\\ \Rightarrow z=\sqrt{2\sqrt 3}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})\\ \Rightarrow \sqrt{2\sqrt 3}\frac{\sqrt 3}{2} +i \sqrt{2\sqrt 3} \frac{1}{2}$
 [/INST] Denesting the radical $\:\sqrt{(1+\sqrt{-3})/2}\:$ can be tackled by employing an easy radical denesting formula that I discovered as a teenager.

Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \  \color{brown}{divide\ out}\ \sqrt{trace} $
Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $  
and, furthermore,  $\rm\:w\:$ has trace $\rm\: =\: w+w' =  (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\:  2\:a$

Here $\ (1+\sqrt{-3})/2\:$ has norm $= 1.\:$ $\rm\ \ \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 1\ $ yields $\ \ (-1+\sqrt{-3})/2\:$
and this has $\rm\:\sqrt{trace}\: =\: \sqrt{-1},\ \ thus\ \ \color{brown}{dividing\ it\ out}\ $ of this gives the sqrt: $\:(\sqrt{3}+\sqrt{-1})/2$
Checking we have
$$\smash[t]{\displaystyle \left(\frac{\sqrt{3}+\sqrt{-1}}{2}\right)^2 =\ \frac{3-1 + 2\sqrt{-3}}{4}\ =\ \frac{1+\sqrt{-3}}2}$$
Therefore $\rm\ \sqrt{\sqrt{3}+3{\it i}}\ =\  \sqrt{2\sqrt 3\left(\dfrac{1+\sqrt{-3}}{2}\right)}\ =\  \sqrt{2}\ 3^{1/4}\:\!\dfrac{(\sqrt{3}+\sqrt{-1})}{2}$
See this answer for general radical denesting algorithms.
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