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
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.
It's not wrong, but it almost certainly not the answer anyone posing such a problem would expect. Typically, one wants an answer 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.
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:
$$e^{ix}=\cos(x)+i\sin(x)$$
So,
$$ \begin{align} 2\cdot 3^{\frac{1}{2}}e^{\frac{3i\pi}{4}}&=2\cdot 3^{\frac{1}{2}}\left(\cos\left(\frac{3\pi}{4}\right)+i \sin\left(\frac{3\pi}{4}\right)\right)\\ &=2\cdot 3^{\frac{1}{2}}\left(-\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right)\\ &=-\frac{2\sqrt{3}}{\sqrt{2}}+\frac{2\sqrt{3}}{\sqrt{2}}i\\ &=-\sqrt{6}+\sqrt{6}i\\ \text{ or }&=(-1+i)\sqrt{6} \end{align} $$
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
