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This question already has an answer here:

I know I can take square root on both sides so I get z and $\sqrt{\sqrt3+3i}$ but is there a way of simplifying $\sqrt{\sqrt3+3i}$ ?

EDIT: Seems like taking the square root may not be the best thing to do (as answer suggests).

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marked as duplicate by Dietrich Burde, Stella Biderman, Leucippus, астон вілла олоф мэллбэрг, C. Falcon Jan 13 '17 at 1:48

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The best way to handle things like this is to write them in polar coordinates, then compute the square root, and then revert back into Cartesian coordinates.

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  • $\begingroup$ Or to use Bill Dubuque's "denesting formula", see the duplicate. $\endgroup$ – Dietrich Burde Jan 12 '17 at 21:14
  • $\begingroup$ I'm curious why this received a down-vote, if whoever did it would like to share. $\endgroup$ – Stella Biderman Jan 12 '17 at 21:43
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Using the polar representation of the complex number $ \sqrt{3}+ 3i$ we have: $$ \sqrt{3}+ 3i=2\sqrt{3}\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i \right)=2\sqrt{3}\left(\cos \left(\frac{\pi}{3} \right)+i\sin \left(\frac{\pi}{3} \right) \right)=2\sqrt{3}e^{i\frac{\pi}{3}} $$

so the values of $z$ that solve the equation are: $$ z=\left[2\sqrt{3}e^{i(\frac{\pi}{3}+2n\pi)} \right]^{\frac{1}{2}}=\sqrt{2\sqrt{3}}e^{i(\frac{\pi}{6}+n\pi)} $$

and the the two values in the interval $[0,2\pi)$ are: $$ z_1=\sqrt{2\sqrt{3}}e^{i\frac{\pi}{6}}=\sqrt{2\sqrt{3}}\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i \right) $$ and $$ z_2=\sqrt{2\sqrt{3}}e^{i\frac{7\pi}{6}}=-\sqrt{2\sqrt{3}}\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i \right) $$

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One way, especially when the argument is not "exact": $$\sqrt{a+bi}=x+yi\Leftrightarrow (x+yi)^2=a+bi\Leftrightarrow x^2-y^2+2xyi=a+bi$$ $$\Leftrightarrow \left \{ \begin{matrix} \displaystyle\begin{aligned} & x^2-y^2=a\\& 2xy=b.\end{aligned}\end{matrix}\right.\Leftrightarrow\ldots$$

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Well, you can use your method, but in complex numbers you'll have to take a small detour, which would entail the following:

Let $z=a+bi$ such that $z^2=9+40i$, then $(a+bi)^2=9+40i$ and therefore

$\begin{align} &a^2+2abi+b^2i^2\\ =&a^2+2abi-b^2\\ =&9+40i \end{align}$

By grouping the real terms, we know that: $\begin{align}a^2-b^2&=9\\ 2abi&=40i \end{align}$

We know that $5^2-4^2=25-16=9$ and we also know that $2*5*4*i=40i$.

Therefore, $z=5+4i$ is a solution.

Can you do the same with your formula?

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  • $\begingroup$ (1) It is not my method, it is an universal and well kown method. (2) Always appear a biquadratic equation, that allows to provide a closed form for the square roots. $\endgroup$ – Fernando Revilla Jan 12 '17 at 22:20
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For square roots of complex numbers, the notation $\sqrt{z}$ should be prohibited and used only for positive real numbers: in this case, it is used to denote the positive square root of the real number. For a complex number there is no canonical way to distinguish between the two square roots of a complex number.

The standard way to find the square roots of a complex number such as $\sqrt 3+3i$ is to use the algebraic form: $\;z=x+iy$, and first identify: $$z^2=x^2-y^2+2ixy=\sqrt 3+3i\iff \begin{cases}x^2-y^2=\sqrt3,\\2xy=3.\end{cases}$$

Second, the trick is to identify the moduli: $\;x^2+y^2=12$. So we have a linear system in $x^2$ and $y^2$: $$ \begin{cases}x^2-y^2=\sqrt3,\\x^2+y^2=\sqrt{12}=2\sqrt 3.\end{cases}\iff \begin{cases}x^2=\dfrac{3\sqrt 3}2,\\ y^2=\dfrac{\sqrt 3}2\end{cases}\iff \begin{cases}x=\pm\frac12\sqrt{6\sqrt3},\\ y=\pm\frac12\sqrt{2\sqrt3}\end{cases}$$ Third, the last condition implies $x$ and $y$ have the same sign, whence the square roots of $\sqrt 3+3i$: $$\pm\frac{\sqrt2\sqrt[4]3}2\bigl(\sqrt3+i\bigr).$$

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