# How can you find the cubed roots of $i$?

I am trying to figure out what the three possibilities of $z$ are such that

$$z^3=i$$

but I am stuck on how to proceed. I tried algebraically but ran into rather tedious polynomials. Could you solve this geometrically? Any help would be greatly appreciated.

-

## 6 Answers

Using Euler's formula, which states $$e^{i \theta} = \cos \theta + i \sin \theta$$ we will see that $$i = 0 + i \cdot 1 = \cos \left( \frac{\pi}{2} + 2n \pi \right) + i \sin \left( \frac{\pi}{2} + 2n \pi \right) = e^{i \left(\frac{ \pi}{2} + 2n \pi \right)}$$ for all integers $n$. Thus, if $z^3 = i$, then $$z = \exp\left[ i \left(\frac{\pi}{6}+\frac{2n\pi}{3}\right)\right]$$ for all integers $n$.

-
Thanks! This makes a lot more sense now. – RXY15 Feb 14 '14 at 2:01

As $\displaystyle i^2=-1,i=-i^3\implies z^3=i=-i^3=(-i)^3\iff z^3-(-i)^3=0$

Now, $\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2),$

$\displaystyle\implies\{z-(-i)\}\{z^2+z(-i)+(-i)^2\}=\{z-(-i)\}(z^2-iz-1)$

If $z-(-i)=0, z=-i$

Else $z^2-iz-1=0\implies z=\dfrac{i\pm\sqrt{i^2-4(-1)}}2=\dfrac{i\pm\sqrt3}2$

-

You can solve this geometrically if you know polar coordinates.

In polar coordinates, multiplication goes $(r_1, \theta_1) \cdot (r_2, \theta_2) = (r_1 \cdot r_2, \theta_1 + \theta_2)$, so cubing goes $(r, \theta)^3 = (r^3, 3\theta)$. The cube roots of $(r, \theta)$ are $\left(\sqrt[3]{r}, \frac{\theta}{3}\right)$, $\left(\sqrt[3]{r}, \frac{\theta+2\pi}{3}\right)$ and $\left(\sqrt[3]{r}, \frac{\theta+4\pi}{3}\right)$ (recall that adding $2\pi$ to the argument doesn't change the number). In other words, to find the cubic roots of a complex number, take the cubic root of the absolute value (the radius) and divide the argument (the angle) by 3.

$i$ is at a right angle from $1$: $i = \left(1, \frac{\pi}{2}\right)$. Graphically:

A cubic root of $i$ is $A = \left(1, \frac{\pi}{6}\right)$. The other two are $B = \left(1, \frac{5\pi}{6}\right)$ and $\left(1, \frac{9\pi}{6}\right) = -i$.

Recalling basic trigonometry, the rectangular coordinates of $A$ are $\left(\cos\frac{\pi}{6}, \sin\frac{\pi}{6}\right)$ (the triangle OMA is rectangle at M). Thus, $A = \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + i\frac{1}{2}$.

If you don't remember the values of $\cos\frac{\pi}{6}$ and $\sin\frac{\pi}{6}$, you can find them using geometry. The triangle $OAi$ has two equal sides $OA$ and $Oi$, so it is isoceles: the angles $OiA$ and $OAi$ are equal. The sum of the angles of the triangle is $\pi$, and we know that the third angle $iOA$ is $\frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$; therefore $OiA = OAi = \dfrac{\pi - \frac{pi}{3}}{2} = \dfrac{\pi}{3}$. So $OAi$ is an equilateral triangle, and the altitude AN is also a median, so N is the midpoint of $[Oi]$: $\sin\frac{\pi}{6} = AM = ON = \frac{1}{2}$. By the Pythagorean theorem, $OM^2 + AM^2 = OA^2 = 1$ so $\cos\frac{\pi}{6} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \dfrac{\sqrt{3}}{2}$.

-

I believe your "polynomial" approach would also have worked, if this is what you meant :

[In this, we are supposing that we knew nothing of the "Euler Identity", DeMoivre's Theorem, or roots of unity, all of which provide quite efficient devices]

If we (probably safely) assume that the solution(s) are complex numbers, and call $\ z \ = \ a + bi \ ,$ with $\ a \$ and $\ b \$ real, we can write the equation as

$$(a + bi)^3 \ = \ a^3 \ + \ 3a^2 b \cdot i \ + \ 3a b^2 \cdot i^2 \ + \ b^3 i^3 \ = \ (a^3 \ - \ 3ab^2) \ + \ (3a^2b \ - \ b^3) \cdot i \ \ = \ \ i \ ,$$

by applying the binomial theorem and "powers of $\ i \$ ". Since the right-hand side of the equation is a pure-imaginary number, this requires that

$$a^3 \ - \ 3ab^2 \ = \ a \ ( a^2 \ - \ 3b^2 ) \ = \ 0 \ \ \text{and} \ \ 3a^2b \ - \ b^3 \ = \ b \ (3a^2 \ - \ b^2) \ = \ 1 \ \ .$$

The first equation presents us with two cases:

I -- $\ a \ = \ 0 \$ :

$$a \ = \ 0 \ \ \Rightarrow \ \ b \ ( \ 0 \ - \ b^2 ) \ = \ -b^3 \ = \ 1 \ \ \Rightarrow \ \ b \ = \ -1 \ \ \Rightarrow \ \ z \ = \ 0 - i \ \ ;$$

II -- $\ a^2 \ - \ 3b^2 \ = \ 0$ :

$$a^2 \ = \ 3b^2 \ \ \Rightarrow \ \ b \ ( \ 3 \cdot [3b^2] \ - \ b^2 \ ) \ = \ 8b^3 \ = \ 1 \ \ \Rightarrow \ \ b \ = \ \frac{1}{2}$$

$$\Rightarrow \ \ a^2 \ = \ 3 \ \left( \frac{1}{2} \right)^2 \ = \ \frac{3}{4} \ \ \Rightarrow \ \ a \ = \ \pm \frac{\sqrt{3}}{2} \ \ \Rightarrow \ \ z \ = \ \frac{\sqrt{3}}{2} + \frac{1}{2}i \ , \ -\frac{\sqrt{3}}{2} + \frac{1}{2}i \ \ .$$

We have found three complex-number solutions to the equation. As Dan says, (one form of) the Fundamental Theorem of Algebra states that this third-degree polynomial with complex coefficients has, in all, three roots (counting multiplicities, which are each 1 here).

We probably wouldn't want to use this method for degrees higher than this, as the algebra would become more difficult to resolve. The techniques described by the other posters are far more generally used.

-
Hmm I guess I made a mistake somewhere when using this method. Thank you for clearing things up! – RXY15 Feb 14 '14 at 2:02

Taking the absolute value of both sides: $|z^3| = |i|$, gives $|z| = 1$. So, $z = \cos (\theta) + i \sin (\theta)$ for some real $\theta$.

Using De Moivre's formula gives $z^3 = \cos(3\theta) + i \sin(3\theta)$. Given that $z^3 = i = 0 + 1i$, this means that $\cos(3\theta) = 0$ and $\sin(3\theta) = 1$. Solving this system gives $3\theta = \frac{\pi}{2} + 2\pi n$, or $\theta = \frac{\pi}{6} + \frac{2 \pi n}{3}$, for any $n \in \mathbb{Z}$.

Plugging in a few values for $n$ gives:

• $n = 0$ → $\theta = \frac{\pi}{6}$ → $z = \frac{\sqrt{3}}{2} + \frac{1}{2} i$
• $n = 1$ → $\theta = \frac{5\pi}{6}$ → $z = \frac{-\sqrt{3}}{2} + \frac{1}{2} i$
• $n = 2$ → $\theta = \frac{3\pi}{2}$ → $z = -i$

And we can stop there because this is a polynomial equation of degree 3, and the Fundamental Theorem of Algebra guarantees that it has at most 3 distinct roots. The solution set is thus $z \in \{ \frac{\sqrt{3}}{2} + \frac{1}{2} i, \frac{-\sqrt{3}}{2} + \frac{1}{2} i, -i \}$.

-

The answer of @Petaro is best, because it suggests how to deal with such questions generally, but here’s another approach to the specific question of what the cube roots of $i$ are.

You know that $(-i)^3=i$, and maybe you know that $\omega=(-1+i\sqrt3)/2$ is a cube root of $1$. So the cube roots of $i$ are the numbers $-i\omega^n$, $n=0,1,2$.

-