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My book (Gamelin ' s Complex Analysis) says that the roots of a complex number are distributed in equal arcs on the circle centered at $0$ with radius $|w|^{1/n} $. Why is the radius centered at zero and why is its radius that length?

Thanks!

Git Gud
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OpieDopee
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2 Answers2

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When you multiply two complex numbers, the result is a new number whose norm (distance from the origin, denoted by |C|) is the product of the original two numbers' norms, and whose angle from the x-axis is the sum of the angles of the original two numbers from the x-axis (great way to remember the trig identities for the sin and cos of the sum of two angles, BTW).

So, "that length" because if |C^(1/n)|^n = |C|, and centered around the origin because every time you multiply the Nth root by itself, you are really rotating around the origin.

This might be bit cryptic because I don't have a good way to draw what is going on...so try graphing the complex number .5 + .5i (which will be on a 45 degree angle from the origin), and then see what happens (graphically) when you square and cube it (you will see the rotation and that the products get closer to the origin), and then try the same for (3^.5)/2 + .5i (this one will rotate as you square and cube it, but the distance from the origin will always be 1), and finally do the same (graph the square and cube) for just 2i.

user146878
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The answer becomes apparent by finding the roots with algebra and then thinking about the roots geometrically. The following six lines I copied from this answer.

For all $n\in \mathbb N, \rho,r\in [0, +\infty[$ and $\theta, \phi\in \mathbb R$ the following folds: $$\begin{align} \left(\rho e^{i\theta}\right)^n=re^{i\phi}&\iff \rho ^ne^{in\theta}=re^{i\phi}\\ &\iff \rho ^n=r\land \cos(n\theta)=\cos(\varphi) \land \sin(n\theta)=\sin(\phi)\\ &\iff \rho ^n=r\land \exists k\in \mathbb Z\left(n\theta=\phi+2k\pi\right).\end{align}$$ Thus, given $w=re^{i\phi}$, the set of its $n$ roots is $\left\{\root n\of r\exp\left(i\dfrac{\phi+2k\pi}{n}\right)\colon k\in \mathbb Z\right\}$ and equals $\left\{\root n\of r\exp\left(i\dfrac{\phi+2k\pi}{n}\right)\colon k\in \{0, \ldots, n-1\}\right\}$.

As you can readily see, the absolute value of the roots of $w$ is $|w|^{1/n}$ and each of the roots is $\dfrac {2k\pi}{n}$ degrees apart.

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Git Gud
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  • What do the larger ^'s mean? – OpieDopee Apr 30 '14 at 16:41
  • @Ethan It's the logical symbol for conjunction. You can get $\land$ by typing $\land$ or $\wedge$. – Git Gud Apr 30 '14 at 16:43
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    I do not understand how this shows the absolute values of the roots of $ w $ are $|w|^{1/n} $... – OpieDopee Apr 30 '14 at 16:44
  • @Ethan Finding the $n$ $n$-roots of $w$ is finding the numbers $z$ such that $z^n=w$. Writing $z=\rho e^{i\theta}$ and $w=re^{i\phi}$ (this can always be done), you just need to follow what I did above to find the roots. Since $|w|=\left|re^{i\phi}\right|=|r|\left|e^{i\phi}\right|=|r|=r$ and similarly $|z|=\rho$, one concludes what you asked. – Git Gud Apr 30 '14 at 16:46