I'd like to find all 3rd roots of this number z = i - 1. Now I've found formulas on how to do it; First we transform the complex number into this form

$$ \sqrt[n]{r} * e^{i\frac {\phi + 2k\pi}{n}} $$ Where n should be 3 (because of 3rd root) and k should be k = n - 1 (Including 0)

Now what is not really clear to me is what to do with this -1? If I transform this I'd pressume I get this.

$$\sqrt[3]{-1} * e^{i\frac {\pi/2 + 2k\pi}{3}}$$

But than what? If it was a 2nd root I could say that it is simply i (the root of a negative number) but since it is a 3rd root I'm not quite sure what to do next?

Now for the phi (angle) I just assumed this; $$i = e^{i \frac {\pi}{2}}$$ And I'm not too sure about this


EDIT: With great help from the people in the comments; I think I've solved it.So as already stated in the comments; $$ r = \sqrt 2, \phi = \frac{3\pi}{4} $$ Now If i would to input that into my formula we should get this;

$$ \sqrt[6]{2} * e^{i\frac{3pi/4 + 2k\pi}{3}} $$

Now as stated before k = 0... n-1;

k = 0 --> $$ \sqrt[6]{2} * e^{i(\pi/4)} $$ k = 1 --> $$ \sqrt[6]{2} * e^{i(11\pi/12)} $$ k = 2 --> $$ \sqrt[6]{2} * e^{i(19\pi/12)} $$

You reckon I did this correctly?


4 Answers 4


Hint: In the formula you used $r$ and $\phi$ are the absolute value and the argument of the complex number $z$.

The absolute value and argument of a complex number $z=x+iy$ can be computed as: $$ r=\sqrt{x^2+y^2}, \quad \phi=\begin {cases}\arccos\frac xr,&y\ge0\\ 2\pi-\arccos\frac xr,&y<0\\ \end {cases}. $$

Particularly for $z=-1+i$ : $$r=\sqrt2,\quad \phi=\frac{3\pi}4. $$

  • $\begingroup$ Okay well that helps a lot; now the only thing left to do is actually input this into the formula? with k = 0... n-1? $\endgroup$
    – codeisfun
    Jan 30, 2021 at 16:00
  • $\begingroup$ @codeisfun Yes, it is. I hope that you correctly understand the expression $\exp\left(i\frac{\phi+2k\pi}n\right)$. $\endgroup$
    – user
    Jan 30, 2021 at 16:21
  • $\begingroup$ Think I got the roots now, but what I would still like to do is to visualise this, maybe get them on a complex plane? $\endgroup$
    – codeisfun
    Jan 30, 2021 at 17:59
  • $\begingroup$ @codeisfun This is a very good and helpful idea! You will find out that given a root of $z$ you can obtain all $n-1$ other roots by the rotation over $\frac{2\pi}n k$ about the origin of the complex plane. $\endgroup$
    – user
    Jan 30, 2021 at 20:59
  • $\begingroup$ I will actually make that in another question,since I'm pretty sure this question has been answered completly.I will try to to do this tommorow,since it is already pretty late where I live. $\endgroup$
    – codeisfun
    Jan 30, 2021 at 23:21

Note that\begin{align}i-1&=-1+i\\&=\sqrt2\left(-\frac1{\sqrt2}+\frac i{\sqrt2}\right)\\&=\sqrt2e^{3\pi i/4}.\end{align}Therefore, the cubic roots of $i-1$ are$$\sqrt[6]2e^{\pi i/4},\ \sqrt[6]2e^{\pi i/4+2\pi i/3}(=\sqrt[6]2e^{11\pi i/12})\text{, and }\sqrt[6]2e^{\pi i/4+4\pi i/3}(=\sqrt[6]2e^{19\pi i/12}).$$


I find these much easier to think about visually. The number you start with is $i-1$, or in the more standard notation $-1+i$. That number is located in the second quadrant of the complex plane. Its modulus (i.e., its absolute value, or distance from $0$) is $\sqrt2$, and its argument (the angle between the positive real axis and a line from $0$ to our point) is $\frac{3\pi}{4}$.

You'll find one cube root by taking the cube root of the modulus, and dividing the argument by $3$. This comes from DeMoivre's theorem on powers of complex numbers. Thus, the cube root of $2^{1/2}$ is $2^{1/6}$, and one third of $\frac{3\pi}{4}$ is $\frac{\pi}{4}$.

Therefore, the principal cube root will be $2^{1/6}$ units from $0$, in the direction of $1+i$. In polar form, it's $2^{1/6}\cdot e^{i\pi/4}$ Considering that the real and imaginary parts are equal, and their squares sum to the square of $2^{1/6}$, we obtain the number $a+bi$, where $a=b=2^{-1/3}$. Writing it with radicals, that's $$\frac{1}{\sqrt[3]{2}}+ \frac{1}{\sqrt[3]{2}}i$$

The other two cube roots will have the same modulus, but their arguments are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ greater than the argument of the principal root.

Does this answer your question?

  • $\begingroup$ Hmm okay,its definetly a step in the right direction.But how do we know the absolute value and argument? I dont know how you figured that out.Still this is all a bit too advanced and confusing to me, we havent covered a lot of these topics in class,but rather the formula I've posted in my question.Is there a way to do it like I intended? $\endgroup$
    – codeisfun
    Jan 30, 2021 at 15:39
  • $\begingroup$ Fair question. Another answer explains how to find the modulus and argument algebraically, although I think seeing the picture is much easier. The number $-1+i$ corresponds to the point $(-1,1)$ on the $xy$-plane. How far is that point from the origin? Well, it lies at the end of the hypotenuse of a right triangle with legs $1$ and $1$, so its distance from $0$ is $\sqrt2$. It lies in the second quadrant, along the line passing through $(-1,1)$, so its angle from the positive horizontal axis is 90 degrees plus 45 degrees, or pi/2 plus pi/4. $\endgroup$ Jan 30, 2021 at 16:25
  • $\begingroup$ To use your formula, you need to first know values for $r$ and $\phi$, and it won't be much help until you have those. In a different answer, you've been shown how to find them algebraically, and I'm telling you how to find them geometrically, since this particular complex number happens to be located where elementary geometry is sufficient. $\endgroup$ Jan 30, 2021 at 16:26


If we let $\omega=e^{i2\pi/3}$, the three roots are

$$\sqrt[3]re^{i\theta/3}, \sqrt[3]re^{i\theta/3}\omega,\sqrt[3]re^{i\theta/3}\omega^2.$$

This is true for any complex number.

Notice that

$$\omega=-\frac12+i\frac{\sqrt3}2,\omega^2=-\frac12-i\frac{\sqrt3}2.$$ The next powers repeat periodically (this is why there are only three distinct roots). The three roots are obtained from any root, multipliying once and twice by $\omega$.

  • $\begingroup$ I would assume it should be $r^{1/3}$ not $r$ in the rhs of the first equation. $\endgroup$
    – user
    Jan 30, 2021 at 20:39
  • $\begingroup$ @user: right. Fixed. $\endgroup$
    – user65203
    Feb 1, 2021 at 8:55

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