The question itself is simple, but I'm weak in math, and I'm training a lot every day to be the best I can, so, working on complex numbers, I got stuck on a simple multiplication:

$\sqrt{3i} * 2i$

I've deduced that the i from $\sqrt{3i}$ is inside the root, so I can't multiply it with the $i$ from $2i$.

I finally got: $2i\sqrt{3i}$ Is that right ? and further, where can I get good material on complex numbers ? thanks in advance

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    $\begingroup$ Sure, by commutativity it's equal to $2i\sqrt{3i}$, but your final answer is likely expected to be in $a+bi$ form. You can split $\sqrt{3i}=\sqrt{3}\sqrt{i}$. To evaluate the square root function of $i$, it would be easiest with polar form. Are you familiar with writing complex numbers in polar form? $\endgroup$ – anon Mar 26 '12 at 0:27
  • $\begingroup$ Whether it's right or not depends on what form of an answer the person asking the question is looking for. Usually, one wants to give an answer in the form $a+bi$ where $a$ and $b$ are real numbers. Do you know how to express $\sqrt i$ in such a form? $\endgroup$ – Gerry Myerson Mar 26 '12 at 0:28
  • $\begingroup$ No, I minded that the both $i$ could be written in form $2\sqrt{2}i^2$ but I guess it's wrong, what would be the right written form ? $\endgroup$ – aajjbb Mar 26 '12 at 0:39

It's not wrong, but it almost certainly not the answer any posing such a problem would expect. Typically, one wants an anser of the form $a + b\mathrm{i}$, where both $a$ and $b$ are real numbers. If you can turn $\sqrt{3\mathrm{i}}$ into this form, multiplying it by $2\mathrm{i}$ will give the desired answer.

To find such numbers, just square one, and see what restrictions you can find: $$ 3 \mathrm{i} = (a+b\mathrm{i})^2 =a^2 - b^2 + 2 a b \mathrm{i} $$

This implies that $a^2 - b^2 = 0$, and thus $a = \pm b$. It also implies that $3 \mathrm{i} = 2 a b \mathrm{i}$, and thus $a b = 3/2$. This means that $a = b$, as $a = -b$ would have no solution over the real numbers. So, $a = b = \pm \sqrt{3/2}$ will give solutions.

For real numbers, there is a convention that when we write the square root symbol, we mean the positive square root. This has to be loosened a bit for complex numbers. The convention is that the principal square root is the one with positive real part. If the real part is zero, it's the one with positive imaginary part. In this case, the standard answer would be $\sqrt{3\mathrm{i}} = \sqrt{3/2} + \sqrt{3/2} \mathrm{i}$.

Multiply this by $2 \mathrm{i}$ and you should have your answer.


This reduces to "denesting" the radical $\:\sqrt{i} = \sqrt{\sqrt{-1}}.\:$ This 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 $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$

So $\:i = \sqrt{-1}\:$ has norm $= 1.\:$ $\rm\:\color{blue}{subtracting\ out}\ \sqrt{norm}\ = \pm1\ $ yields $\ i\pm 1$

with $\rm\:\sqrt{trace}\: =\: \sqrt{trace({\it i}\pm 1)}\ =\ \sqrt{\pm2},\ \ so\ \ \color{brown}{dividing\ it\ out}\ $ yields $\smash{\ \dfrac{1\pm i}{\sqrt{\pm2}}\:= \dfrac{\pm(1+{\it i})}{\sqrt{2}}}$

Indeed, checking we have $$\displaystyle \left(\dfrac{1\pm i}{\sqrt{\pm2}}\right)^2\ =\ \frac{1 \pm 2\:\!i -1}{\pm2}\ =\ i$$

Below is another example from a prior question.

Note that $\ 3 + 2\sqrt{2}\ $ has norm $\ (3+2\sqrt{2})\:(3-2\sqrt{2})\ =\ 9 - 4\cdot 2\ =\ 1$

So $\rm\:\color{blue}{subtracting\ out}\ \sqrt{norm}\ = \pm1\ $ yields $\rm\ 3 + 2\sqrt{2}\: -\: \pm1\ =\ 2 + 2\sqrt{2}\ \ or\ \ 4 + 2\sqrt{2} $

$\rm\qquad \sqrt{trace(2+2\sqrt{2})}\ =\ \sqrt{4}\ =\ 2,\quad\ \ so\ \quad\ \color{brown}{dividing\ it\ out}\ \ \ (2+2\sqrt{2})/2\ =\ 1+\sqrt{2}$

$\rm\qquad \sqrt{trace(4+2\sqrt{2})}\ =\ \sqrt{8}\ =\ 2\sqrt{2}\quad so\quad \color{brown}{dividing\ it\ out}\ \ \ (4+2\sqrt{2})/(2\sqrt{2})\: =\: \sqrt{2}+1$

Indeed $\rm\ (1 + \sqrt{2})^2\ =\ 3 + 2\sqrt{2}\:.\ $ It is an easy exercise to check that the formula is correct.

With experience from a few worked examples, one can swiftly mentally denest such radicals.

  • 1
    $\begingroup$ Note that this method assumes $\sqrt{\mathrm{norm}} \in \mathbb Z$. $\endgroup$ – Ben Frankel Oct 25 '14 at 8:05
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    $\begingroup$ @Ben If not then a denesting does not exist. $\endgroup$ – Bill Dubuque Dec 18 '14 at 4:09

Here's how I did it:

$$ \begin{align} \sqrt{3i}\cdot 2i&=(3i)^{\frac{1}{2}}\cdot 2i\\ &=3^{\frac{1}{2}}i^{\frac{1}{2}}\cdot 2i\\ &=2\cdot 3^{\frac{1}{2}}i^{\frac{3}{2}}\\ &=2\cdot 3^{\frac{1}{2}}(-1^{\frac{1}{2}})^{\frac{3}{2}}\\ &=2\cdot 3^{\frac{1}{2}}(-1)^{\frac{3}{4}} \end{align} $$

Using the identity of Euler, one can write this more appropriately: $e^{i\pi }=-1$

$$2\cdot 3^{\frac{1}{2}}(-1)^{\frac{3}{4}}=2\cdot 3^{\frac{1}{2}}e^{\frac{3i\pi}{4}}$$

If you wish to be a bit out there, you can recall de Moivre:



$$ \begin{align} 2\cdot 3^{\frac{1}{2}}e^{\frac{3i\pi}{4}}&=2\cdot 3^{\frac{1}{2}}\left(\cos\left(\frac{3\pi}{4}\right)+i \sin\left(\frac{3\pi}{4}\right)\right)\\ &=2\cdot 3^{\frac{1}{2}}\left(-\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right)\\ &=-\frac{2\sqrt{3}}{\sqrt{2}}+\frac{2\sqrt{3}}{\sqrt{2}}i\\ &=-\sqrt{6}+\sqrt{6}i\\ \text{ or }&=(-1+i)\sqrt{6} \end{align} $$


I would solve this problem like this:

$$\sqrt{i} = \exp (\frac {\log i}{2}) = \exp (\frac {\log |i| + i\arg i }{2}) = \exp (\frac {\log 1 + i\pi/2 }{2}) = \exp (\frac {i\pi }{4}) = \cos (\pi/4) + i\sin (\pi/4) = \frac {1}{\sqrt{2}} + i\frac {1}{\sqrt{2}}$$ Thus we have (together):

$$\sqrt{i3}*i2=i2\sqrt{3}(\frac {1}{\sqrt{2}} + i\frac {1}{\sqrt{2}})$$ Expanding, we have:

$$i2\sqrt{3}(\frac {1}{\sqrt{2}} + i\frac {1}{\sqrt{2}}) = i\frac{2\sqrt{3}}{\sqrt{2}}-\frac{2\sqrt{3}}{\sqrt{2}} = 2\sqrt{\frac{3}{2}}(i-1)=\sqrt{6}(i-1)$$

You could modify this to whatever form.


Every complex number has 2 different square roots so you will get two different answers. Start with this wikipedia article


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