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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.

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At once, you have $2\sqrt[3]{i}=2(-1)^\frac16$. You now have to pick which root of $x^6+1=(x^2+1)(x^4-x^2+1)$ is needed... – J. M. Dec 21 '11 at 6:04
Regarding J.M.'s comment, note that there are only 3 cube roots of $8i$, whereas there are 6 sixth roots of $-1$. Do not take the equation $2\sqrt[3]{i}=2(-1)^{\frac{1}{6}}$ seriously. It is true that if $z^3=8i$, then $(z/2)^6=-1$, but the reverse implication does not hold. (I'm sure J.M. is aware of this, but I was afraid it could be misleading to the OP or others who aren't aware.) – Jonas Meyer Dec 21 '11 at 6:41
Yes, Jonas has a point. Since $6$ is even, three of the roots of $x^6+1$ are extraneous for the task at hand. (Similar care is needed when manipulating equations with radicals, anyway.) – J. M. Dec 21 '11 at 7:33
up vote 6 down vote accepted

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,

$$ 8i = 8 e^{i\pi/2} \ . $$


$$ \sqrt[3]{8e^{i\pi/2}} = \sqrt[3]{8} e^{i(\pi/2 + 2k\pi)/3} \qquad \text{for} \quad k=0,1,2 \ . $$

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:

$$ 2e^{i\pi/6} = 2\left(\cos\frac{\pi}{6} + i \sin\frac{\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} + i\frac{1}{2} \right) \ , $$

$$ 2e^{i5\pi/6} = 2\left(\cos\frac{5\pi}{6} + i \sin\frac{5\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} - i\frac{1}{2} \right) \ , $$


$$ 2e^{i3\pi/2} = 2\left(\cos\frac{3\pi}{2} + i \sin\frac{3\pi}{2}\right) = -2i \ . $$

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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 $2$.

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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 $-\sqrt{3} + i$.

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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 $\cos150^\circ+i\sin150^\circ= -\sqrt{3}/2 + i/2$.)

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Multiplying by $\cos30^\circ+i\sin30^\circ$ only rotates by $30^\circ$ counterclockwise. When you include the multiplication by $2$, the three cube roots of $8i$ work out to be $\sqrt{3} + i$, $-\sqrt{3}+i$ and $-2i$. – Dilip Sarwate Dec 21 '11 at 15:52

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