How to find $\sqrt[3]{8i}$ How do I find the following cube root?
$$\sqrt[3]{8i} = ?$$
I tried by adding $\sqrt[3]{i^3 + 8i + i}$ but that is where my imagination quits.
 A: If you want a general method for finding roots of complex numbers, you may try my answer here, for instance.
In your case, it would work as follows: write your complex number in exponential form,
$$
8i = 8 e^{i\pi/2} \ .
$$
Then, 
$$
\sqrt[3]{8e^{i\pi/2}} = \sqrt[3]{8} e^{i(\pi/2 + 2k\pi)/3} \qquad \text{for} \quad k=0,1,2 \ .
$$ 
That is,
$$
\sqrt[3]{8e^{i\pi/2}} = 2 e^{i(\pi/6 + 2k\pi/3)} \qquad \text{for} \quad k=0,1,2
$$
And this gives:
$$
2e^{i\pi/6} = 2\left(\cos\frac{\pi}{6} + i \sin\frac{\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} + i\frac{1}{2} \right) \ ,
$$
$$
2e^{i5\pi/6} = 2\left(\cos\frac{5\pi}{6} + i \sin\frac{5\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} - i\frac{1}{2} \right) \ ,
$$
and
$$
2e^{i3\pi/2} = 2\left(\cos\frac{3\pi}{2} + i \sin\frac{3\pi}{2}\right) = -2i \ .
$$
A: For this particular problem, we can find one cube root "by inspection." We can try $2i$, its cube is $-8i$. The fix is easy: $-2i$ works. Then to list all the cube roots of $8i$, we multiply the one we found by all the cube roots of $1$, that is, the solutions of $x^3=1$. These are easy to find, since $x^3-1=(x-1)(x^2+x+1)$.   So we multiply $-2i$ by $1$, also by $\frac{-1+i\sqrt{3}}{2}$, also by  $\frac{-1-i\sqrt{3}}{2}$.
But this is not the right general approach.  What we should do in general is to first express our number $8i$ as $re^{i\theta}$, or equivalently as $r(\cos\theta+i\sin\theta)$.  Then one cube root is $r^{1/3}e^{i\theta/3}$. The other cube roots can be obtained by multiplying by the cube roots of unity other than $1$.
In our particular example, $r=8$.  We want $\cos\theta+i\sin\theta=i$. So $\theta=\pi/2$ will do the job. Thus one cube root is $2(\cos(\pi/6)+i\sin(\pi/6))$. The others can be obtained by multiplying by cube roots of unity. Alternately, use also $\theta=\pi/2+2\pi$, $\theta=\pi/2+4\pi$, and divide each by $3$. Or else equivalently write that the roots are $2e^{i(\pi/2+2k\pi)/3}$, where $k=0$, $1$, or $2$.   
A: Here is another way that avoids exponential functions 
for this problem but is not a method that I would 
recommend in general, e.g. to find $\sqrt[10]{8i}$,
for which the method described Andre Nicolas's and
Agusti Roig's answers work much better.
Suppose $a$ and $b$ are real numbers such that $(a+bi)^3 = 8i$.  We have
$$\begin{align*}
(a+bi)^3 &= a^3 + 3a^2bi + 3a(bi)^2 + (bi)^3\\
&= (a^3 -3ab^2) + (3a^2b - b^3)i\\
&= 0 + 8i,
\end{align*}$$
and so
$$\begin{align*}
a^3 - 3ab^2 = a(a^2 - 3b^2) &= 0\\
3a^2b - b^3 = b(3a^2 - b^2) &= 8
\end{align*}$$


*

*From the first equation we see that one possible value for $a$ is $0$,
and in this case, the second equation reduces to $-b^3 = 8$ giving $b = -2$.
In other words, one cube root of $8i$ is $-2i$.

*Alternatively, if $a \neq 0$, then from the first equation we see 
that it must be that  $a^2 - 3b^2 = 0$
so that $a = \pm \sqrt{3}b$.  Substituting into the second equation, we 
get  $b(9b^2 - b^2) = 8b^3 = 8$, that is $b^3 = 1$. Now, $b = 1$ is 
the only real number solution to $b^3 = 1$, and so we get that 
the other two
cube roots of $8i$ are $\sqrt{3} + i$ and $-\sqrt{3} + i$. 
A: Multiplication by $8i$ consists of making complex numbers $8$ times as far from $0$ as they were and rotating $90^\circ$ counterclockwise.  (Remember that multiplying by $i$ consists of rotating $90^\circ$ counterclockwise.)
So you want to multiply by something three times and have that amount to multiplying by $8i$.
Therefore: Multiply by $2$ and rotate only $30^\circ$ counterclockwise.  That gives you
$$
\cos30^\circ+i\sin30^\circ = \frac{\sqrt{3}}{2} + \frac12 i.
$$
The other two cube roots come from rotating that through $1/3$ circle, i.e. $120^\circ$, and $2/3$ circle, i.e. $240^\circ$.  Since $30+240=270$ is mod-$360$ congruent to $-90$, one of the cube roots is at $-90^\circ$, so pointing straight downward in the usual depiction of the plane.  In other words, it is $-2i$.  (And the third is $\cos150^\circ+i\sin150^\circ= -\sqrt{3}/2 + i/2$.)
A: Look this site 
http://fatosmatematicos.blogspot.com/2011/09/identidade-de-euler-e-as-raizes.html
