# 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: $w=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})=z^2\\ \Rightarrow z=\sqrt{2\sqrt 3}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})\\ \Rightarrow \sqrt{2\sqrt 3}\frac{\sqrt 3}{2} +i \sqrt{2\sqrt 3} \frac{1}{2}$

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It's factorization, if you multiply the parenthesis by $2 \sqrt 3$ you get what you had at first and for the second part, $\cos \pi/3 =1/2$ isn't it? –  Gigili May 23 '12 at 17:38
Let $z^2 = w$ is from the definition of the square root. Really, they are saying: we want to find a $z$ so that $z^2 = w$. –  Ronald May 23 '12 at 17:45
Thanks for your help, understood it (better) now! –  Derrick May 23 '12 at 18:02

Given a complex number $z = a + i b$, its absolute value $|z| = \sqrt{a^2+b^2}$, naturally the quotient $\frac{z}{|z|}$ has unit absolutely value, hence $\frac{z}{|z|} = \mathrm{e}^{i \theta} = \cos(\theta) + i \sin(\theta)$ for some angle $\theta$.

In the case at $a=\sqrt{3}$ and $b=3$, thus $\sqrt{a^2+b^2} = \sqrt{3+3^2} = 2 \sqrt{3}$. Therefore $\frac{z}{|z|} = \frac{\sqrt{3}}{2 \sqrt{3}} + i \frac{3}{2 \sqrt{3}} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$. Solving for $\cos(\theta) = \frac{1}{2}$ and $\sin(\theta) = \frac{\sqrt{3}}{2}$ for $0 \leqslant \theta < 2\pi$ gives $\theta = \frac{\pi}{3}$.

Finding the square root proceeds as follows. Let $w = |w| \mathrm{e}^{i \phi}$, then $$|w|^2 \mathrm{e}^{2 i \phi} = 2 \sqrt{3} \mathrm{e}^{i \pi/3}$$ Taking the absolute value we must have $|w|^2 = 2 \sqrt{3}$, hence $|w| = \sqrt{2} 3^{1/4}$. When solving for the angle $\phi$, remember that there are two roots for $0 \leqslant \phi <2 \pi$.

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Thanks @Sasha, may I ask, is $\frac{z}{|z|} = \mathrm{e}^{i \theta}$ a standard result that one can quote/use? –  Derrick May 23 '12 at 18:04
Yes. Equation for unit absolute value is $a^2+b^2 = 1$, which can be explicitly solved as $a = \cos(\theta)$ and $b = \sin(\theta)$ for some $0 \leqslant \theta < 2 \pi$. Then Euler's formula allows to write $\cos(\theta) + i \sin(\theta) = \mathrm{e}^{i \theta}$. –  Sasha May 23 '12 at 18:07
Ok, thanks for your help again! –  Derrick May 23 '12 at 18:16
@Derrick Yes, except two solutions are possible for $\phi$, $\phi_1=\frac{\pi}{6}$ and $\phi_2 = \frac{\pi}{6} + \pi$. This will give $z_2 = -z_1$. –  Sasha May 23 '12 at 19:00
@Derrick Yes, you can verify by squaring that both $z_1 = \sqrt{2\sqrt 3}\frac{\sqrt 3}{2} +i \sqrt{2\sqrt 3} \frac{1}{2}$ and $z_2 = -\sqrt{2\sqrt 3}\frac{\sqrt 3}{2} - i \sqrt{2\sqrt 3} \frac{1}{2}$ (both terms have negative sign), verify $z_1^2 = z_2^2 = \sqrt{3} + 3 i$. –  Sasha May 23 '12 at 21:36

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 $\rm\ \sqrt{\sqrt{3}+3{\it i}}\ =\ \sqrt{2\sqrt 3\left(\dfrac{1+\sqrt{-3}}{2}\right)}\ =\ \sqrt{2}\ 3^{1/4}\:\!\dfrac{(\sqrt{3}+\sqrt{-1})}{2}$

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+1 Nice trick. Can it be generalized beyond surds? –  Sasha May 23 '12 at 18:50
Thhanks for this great tip! –  Derrick May 23 '12 at 18:59
@Sasha I vaguely recall that I did find some higher degree generalizations. But that was almost a few decades ago so, alas, I don't recall the details. You might be able to rediscover them (it might help to know some Kummer-Galois theory - see the above link). –  Bill Dubuque May 23 '12 at 19:04

The first step is to find the modulus-argument form: the modulus of the complex number comes from $\sqrt{x^2 + y^2}$.

They have taken this as a factor, leaving an 'obvious' way to find the argument by knowing the trigonometric exact values. It's perhaps easier to find the argument using arctan with reference to a picture of the Argand diagram.

Let $z^2 = w$ is from the definition of the square root. Really, they are saying: we want to find a $z$ so that $z^2 = w$.

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Thanks Ronald, I understand it better now. Obviously I guess it required some intuition to get take out the factor "well". –  Derrick May 23 '12 at 18:03

The general procedure (for square roots) goes as follows.

We want the square roots of $a+ib$ (where at least one of $a$ or $b$ is non-zero). Note that $$a+ib=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}+i\frac{b}{\sqrt{a^2+b^2}}i\right).\tag{1}$$

For brevity, write $x=\frac{a}{\sqrt{a^2+b^2}}$ and $y=\frac{b}{\sqrt{a^2+b^2}}$. Since $x^2+y^2=1$, there is a unique $\theta$ in the interval $[0,2\pi)$ such that $x=\cos\theta$ and $y=\sin\theta$.

In your example, $x=\frac{1}{2}$ and $y=\frac{\sqrt{3}}{2}$, so, from "special angles" we recognize that $\theta=\pi/3$.

Now the square roots of $\cos\theta+i\sin\theta$ are $\pm(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})$, and from this we find that the square roots of $a+bi$ are $$\pm\sqrt[4]{a^2+b^2}\left(\cos\tfrac{\theta}{2}+i\sin\tfrac{\theta}{2}\right).$$

Added: In language close to the language used in the post, let $w=a+bi$. We want to solve the equation $z^2=w$. Let $z=r(\cos\phi+i\sin\phi)$, where $r$ is positive. Then $z^2=r^2(\cos 2\phi+i\sin 2\phi)$. Comparing with the expression $(1)$, we find that $r^2=\sqrt{a^2+b^2}$ and therefore $r=\sqrt[4]{a^2+b^2}$.

Also, we want $\cos2\phi +i\sin 2\phi=\cos\theta+i\sin\theta$. Obviously $\phi=\frac{\theta}{2}$ works. But also, more subtly, so does $\phi=\frac{\theta}{2}+\pi$, which gives you the negative of the square root picked out by $\phi=\frac{\theta}{2}$.

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Thanks @andrenicolas , may I ask, how did you get that $a+ib=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}+i\frac{b}{\sqrt{a^2+b^2}}i\ri‌​ght)$? –  Derrick May 23 '12 at 18:07
@Derrick: The correctness is easy, we just multiply through. If your question is why I wrote it that way, it is because I wanted to express $a+ib$ as a positive real times a number $x+iy$ with $x^2+y^2=1$, so that I could use the De Moivre Theorem to take the square root. But the right answer to your question is that I remembered that it is a strategy that works. –  André Nicolas May 23 '12 at 18:13
Ok, thanks for your help, I guess I need to work/learn these strategies! –  Derrick May 23 '12 at 18:15
@Derrick: Minus signs in both places (or else use parentheses, $-(u+iv)$). –  André Nicolas May 23 '12 at 20:49
There is an extra $i$ in equation (1), which I'm sure is not intended. Unfortunately I cannot edit it to remove only one character! –  Ronald May 23 '12 at 21:36