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As the title suggests, I am having trouble finding the cube roots of $-i$. I looked at the similar post of finding the cube roots of just $i$ but I am having some confusion.

Complex numbers are of the form $z=a+bi$.

$$-i= 0+(-1)i$$

Giving a modulus of $|z|=\sqrt{0^2+(-1)^2}=1$

Then finding $\arg(-1)=\tan^{-1}\left(\large\frac{b}{a}\right)$, right? But our $a=0$, so how can we continue from here?

I'm sure there is some small detail I am overlooking. Thanks in advance.

Michael Hardy
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Vincent
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  • Look to $(0,-i)$ in the Argand-Gauss plane, what is the angle? (The angle, of course, is considered from $0x$-axis). – DiegoMath Oct 04 '14 at 03:02
  • Use the two-argument arctangent function atan2 to find the angle. https://en.wikipedia.org/wiki/Argument_(complex_analysis)#Computation – Steve Kass Oct 04 '14 at 04:26
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    Draw a picture, and deduce the argument from that. Feel sorry for the computer program that cannot "see" it, and is in desperate need for a formula. – Jyrki Lahtonen Oct 04 '14 at 07:49

4 Answers4

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You can say $\tan^{-1}(1/0)=\tan^{-1}(\infty)= 90^\circ$ (or you can say $\pi/2$ radians), and this "$\infty$" is neither $+\infty$ nor $-\infty$, but rather is the single $\infty$ that is at both ends of the real line. That is the "$\infty$" that is appropriate when talking about the ranges of trigonometric functions or about either the domains or the ranges of rational functions, and it makes all such functions continuous at every point in their domains.

So a cube root is $\cos30^\circ+i\sin30^\circ$ (or $\pi/6$ radians if you like).

(Then you also get the two other cube roots that come from adding $\pm360^\circ/3=\pm120^\circ$.)

PS: I had read the question is asking for cube roots of $i$. I now see that it now says $-i$. With $-i$, just take the complex conjugates of the numbers above. Or, working from scratch: for $-i$, we need the other angle whose tangent is $\infty$, namely $-90^\circ$. The answer is then $\cos(-30^\circ)+i\sin(-30^\circ)$, which is the same as $\cos30^\circ-i\sin30^\circ$, and then add $\pm120^\circ$ to get the other two cube roots.

Michael Hardy
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  • So, $$\cos \frac{\pi}{6}+i\sin \frac{\pi}{6}$$

    $$\cos \frac{5\pi}{6}+i\sin \frac{5\pi}{6}$$

    $$\cos \frac{3\pi}{2}+i\sin \frac{3\pi}{2}$$?

     – Vincent Oct 04 '14 at 03:19
  • Yes. And you should be able to check those by writing $(a+b)^3=a^3+3a^2b+3ab^2+b^3$, etc. ${}\qquad{}$ – Michael Hardy Oct 04 '14 at 03:24
  • None of these is a cube root of $-i$. – Did Oct 04 '14 at 07:44
  • Uh oh. What did we do wrong? – Vincent Oct 04 '14 at 07:51
  • I did write them as: $\pm \large\frac{\sqrt{3}+i}{2}$ and $-i$, if that was the problem. – Vincent Oct 04 '14 at 08:09
  • Vincent: You might want to compute from scratch $(-i)^3$ and to compare the result with $-i$. If indeed $(-i)^3=-i$, then I turned senile and should stop posting comments on math.SE, if $(-i)^3\ne -i$, then $-i$ is not a cube rot of $-i$ and something needs to be modified in this post. – Did Oct 04 '14 at 08:15
  • @Did, I agree, the OP has accepted an incorrect answer, which computes the cube root of $i$, not $-i$. Both he and Michael Hardy should reexamine this solution, to see where it went wrong. – Barry Cipra Oct 04 '14 at 13:53
  • I had read the question as asking for the cube root of $i$. For $-i$, we need the other angle whose tangent is $\infty$, namely $-90^\circ$. The answer is then $\cos(-30^\circ)+i\sin(-30^\circ)$, which is the same as $\cos30^\circ-i\sin30^\circ$, and then add $\pm120^\circ$ to get the other two cube roots. – Michael Hardy Oct 04 '14 at 16:08
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There's a simpler and more elementary way of determining the cube roots of $i$.

We have

$$a+bi=\sqrt[3]{-i}$$

Cubing both sides give

$$a^3+3a^2bi-3ab^2-b^3i=-i$$

so

$$a^3-3ab^2=0$$ $$3a^2b-b^3=-1$$

And from there it's easy solving the system.

Edward Jiang
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  • "Dividing the second by $b$..." What happened to the $-1$ on the other side? – abiessu Oct 04 '14 at 03:12
  • Also, with that problem in the second equation it appears that the original question still stands, and we have now just moved the problem to a different variable. – abiessu Oct 04 '14 at 03:15
  • @abiessu, no, it's OK as it currently stands, because the variables $a$ and $b$ are real. The first equation implies either $a=0$ or $a^2=3b^2$. When you plug these into the second you get either $b^3=1$ or $8b^3=-1$. Because $b$ can only be real, these easily solve to $b=1$ and $b=-1/2$, and the second of these plugs back in to give $a=\pm\sqrt3/2$. – Barry Cipra Oct 04 '14 at 03:37
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    I don't agree that this is simpler or more elementary. ${}\qquad{}$ – Michael Hardy Oct 04 '14 at 03:43
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    It's easier for, say, a high school student to understand this method. – Edward Jiang Oct 04 '14 at 03:45
  • I do agree that it is more elementary, a person could figure this out on their own without knowing Euler's Formula or DeMoivre's Formula, but it's unlikely that a person would figure either of those out on their own. – DanielV Oct 04 '14 at 04:26
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Here's a generalized formula for powers of $i$, taken directly from Euler's formula:

\begin{align} i &= e^{i\pi \over 2}\\ e^{ix} &= \cos x + i\sin x \end{align}

Combining the two, we get:

$$ i^x = \left(e^{i\pi \over 2}\right)^x = e^{i{\pi x \over 2}} $$

Thus:

$$ i^x = \cos\left(\pi x \over 2\right) + i\sin\left(\pi x \over 2\right) $$

In this specific case, we see that one cube root of $i$ is given by:

$$ i^{1 \over 3} = \cos{\pi \over 6} + i \sin{\pi \over 6} = {\sqrt 3 + i \over 2} $$

As was mentioned, the argument here is $\pi \over 6$. You can find the remaining two cube roots by adding and subtracting $2\pi \over 3$ from the argument.

  • Actually no $i^x$ for $x$ noninteger (such as $x=1/3$ used later in the answer) is a well defined complex number--which explains why the notation is seldom used except in lazy textbooks and/or popularization tracts and/or math.SE answers. – Did Oct 04 '14 at 07:46
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You can find one solution by inspection: $i^3=(i^2)i=-i$.

So $z=i$ is one solution of the equation $z^3+i=0$.

If you happen to know the cube roots of $1$, you can find all three cube roots of $-i$ by multiplying each of the cube roots of $1$ by $i$:

$i\cdot1=i,\ \ i\cdot(-\frac12+\frac{\sqrt3}2i)=-\frac{\sqrt3}2-\frac12i,\ \ i\cdot(-\frac12-\frac{\sqrt3}2i)=\frac{\sqrt3}2-\frac12i$.

Alternatively, since $z-i$ is a factor of $z^3+i$, you can use long division to find the factorization $z^3+i=(z-i)(z^2+iz-1$) and then use the quadratic formula to solve the quadratic equation $z^2+iz-1=0$.

bof
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