$z^2 = \sqrt3+ 3i$ (complex equation) [duplicate]

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).

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

• – Ng Chung Tak Jan 12 '17 at 23:09

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.

• Or to use Bill Dubuque's "denesting formula", see the duplicate. – Dietrich Burde Jan 12 '17 at 21:14
• I'm curious why this received a down-vote, if whoever did it would like to share. – Stella Biderman Jan 12 '17 at 21:43

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 $$z_2=\sqrt{2\sqrt{3}}e^{i\frac{7\pi}{6}}=-\sqrt{2\sqrt{3}}\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i \right)$$

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

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?

• (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. – Fernando Revilla Jan 12 '17 at 22:20

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 $\sqrt 3+3i$: $$\pm\frac{\sqrt2\sqrt3}2\bigl(\sqrt3+i\bigr).$$