# What are the three cube roots of $1-i$?

I am trying to find the three cube roots of $1-i$ and I am unsure of how to proceed. My math teacher recommended I use de Moivre's formula, however I do not know how to set it up. How do you suggest I continue?

de Moivre's formula

$$(\cos x + i\sin x)^n = \cos(nx) + i\sin(nx)$$

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Work exponentially: Write $1-i$ in the polar form $z=1-i=re^{i\theta}$. I leave it to you to find $r$ and $\theta$. Now, assume $w=se^{i\rho}$ satisfies $w^3=z$. Then, by ordinary rules of exponentiation, you have that $z=w^3=s^3e^{i3\rho}$. This results in the equality $re^{i\theta}=s^3e^{i3\rho }$. So, you get two equations: $r=s^3$ and $e^{i\theta}=e^{i3\rho}$. Solving the first (which is just good old real numbers) gives you the unique solution $s$ is the cubed root of $r$. The second equality means that $\theta$ and $3\rho$ differ by a multiple of $2\pi$ so that $\rho =1/3 \cdot (\theta + 2\pi k)$, $k=0,1,2$, giving you three unique solutions.

This is a general method that will work for extracting any root from any complex number. The same technique will work for computing logarithms. It's completely algebraic which is good and bad. It's completely algorithmic and fail safe but misses out on the geometric aspect.

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It should be noted that Jack didn't mention the exponential version of polar form, but $e^{i\theta}=\cos\theta+i\sin\theta$ connects this with the question. – Mark S. Feb 8 '13 at 22:55

Another way without using DeMoivre's formula that works for cube roots is set

$x=1$ and $y=-i$ and then find any real roots of the following equation

$$\frac{(-64a^9+(48x)a^6+((15(x)^2)-3(3y)^2)a^3+(x)^3)}{-64} = 0$$

This expands to $$a^9 - \frac{3}{4}a^6 - \frac{21}{32}a^3 - \frac{1}{64} = 0$$

and has three real roots $a_1=\frac{1}{4}(\sqrt[6]{432}+\sqrt[6]{16}); $$a_2=\frac{1}{4}(-\sqrt[6]{432}+\sqrt[6]{16}); a_3=\frac{1}{2}(-\sqrt[6]{16}). Next solve the equations$$a_1^3+3a_1b_1^2 = 1a_2^3+3a_2b_2^2 = 1a_3^3+3a_3b_3^2 = 1$$(where 1 is the value set as x) for b_1, b_2 and b_3, or b_1=\pm\frac{1}{4}(\sqrt[6]{432}-\sqrt[6]{16})i; b_2=\pm\frac{1}{4}(\sqrt[6]{432}+\sqrt[6]{16})i;$$b_3=\pm\frac{1}{2}(\sqrt[6]{16})i$ . So the three cube roots of $$\sqrt[3]{1-i}$$ are

$$a_1+b_1=\frac{1}{4}(\sqrt[6]{432}+\sqrt[6]{16})+\frac{1}{4}(-\sqrt[6]{432}+\sqrt[6]{16})i$$

$$a_2+b_2=\frac{1}{4}(-\sqrt[6]{432}+\sqrt[6]{16})+\frac{1}{4}(\sqrt[6]{432}+\sqrt[6]{16})i$$

$$a_3+b_3=\frac{1}{2}(-\sqrt[6]{16})+\frac{1}{2}(-\sqrt[6]{16})i$$

Combinations of $a_1$, $a_2$ and $a_3$ and $b_1$, $b_2$ and $b_3$ will also give the three roots of $\sqrt[3]{1+i}$.

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Let $\zeta_n$ denote the first $n$-th root of unity counterclockwise from $1$.

If you're familiar with your roots of unity (or think about it graphically), you'll recall that

$$\zeta_8 = \frac{1 + \mathbf{i}}{\sqrt{2}}$$

and so you can immediately determine from its graphical position in the plane

$$1 - \mathbf{i} = \sqrt{2} \zeta_8^7$$

If you know or can intuit some group theory, you'll realize that because $\zeta_8^8 = 1$ and because $3$ and $8$ are relatively prime, I should be able to rewrite $\zeta_8^7$ as something with an exponent divisible by 3. e.g.

$$\zeta_8^7 = \zeta_8^{15}$$

At this point, the third roots of $1 - \mathbf{i}$ are easy to determine: they are

$$\sqrt[6]2 \zeta_8^5 \qquad \qquad \sqrt[6]2 \zeta_8^5 \zeta_3 \qquad \qquad \sqrt[6]2 \zeta_8^5 \zeta_3^2$$

where

$$\zeta_3 = \frac{-1 + \sqrt{3} \mathbf{i}}{2}$$

If I was so inclined, I could rewrite these in terms of $\zeta_{24}$, because $\zeta_8 = \zeta_{24}^3$ and $\zeta_3 = \zeta_{24}^8$.

If the latter two roots of $1 - \mathbf{i}$ aren't obvious, recall that all of the $n$-th roots of a number differ by $n$-th roots of unity; that is, if $y$ is an $n$-th root of $x$, then so is $y \zeta_n$. This is easy to check directly by taking the $n$-th power.

$$(y \zeta_n)^n = y^n \zeta_n^n = y^n = x$$

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Once you know the angles for the roots, you can find them explicitly in radicals by using the half angle formulas on two of the roots (the one at -15∘, for instance). The third root doesn't even require that.

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Express $1-i$ in polar form:

$$1-i=\sqrt{2} e^{-i \pi/4}$$

So that

$$(1-i)^{1/3} = 2^{1/6} e^{-i \pi/12 + i k 2\pi/3}$$

Where $k \in \mathbb{Z}$. So to find three unique roots, choose, $k \in \{0,1,2\}$.

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Draw it on coordinate system, then pick the third of its angle closed with the right wing of $x$-axis (containing $1$): that is, now $-45^\circ/3$. Then imagine the regular triangle in the circle around $0$ with this one vertex at $-15^\circ$.

The length (absolute value, distance from $0$) gets its cubic root: $$(r(\cos\phi+i\sin\phi))^3 = r^3(\cos(3\phi)+i\sin(3\phi))$$

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