1

Let $z^n=w$ where $w$ is a complex number. There will be $n$ solutions for this equation. Let the solutions be $z_1,z_2,\ldots,z_{n-1},z_n$.

If $arg(z_2)-arg(z_1)=d$, it is my observation that $arg(z_x)=arg(z_{x-1})+d$ where $x \in \mathbb{Z}$.

Further, plotting the solutions on an argand diagram divides the diagram into equally sized portions. For instance, if n=3, the plotted solutions will divide the diagram into 3 equal sections.

Is there some intuitive explanation as to why:

1) The solutions to $z^n=w$ form an arithmetic sequence?

2) When plotted, the solutions divide up the argand diagram into $n$ sections of equal size?

Haotian H.
  • 41

2 Answers2

3

Let $w = re^{i\theta}$ Now, $z=w^{\frac{1}{n}} = r^\frac{1}{n}e^{i\frac{\theta}{n}}$

By De Moivre's theorem, the solutions will be \begin{equation} z = r^{\frac{1}{n}}\Bigg(\cos{\Big(\frac{2k\pi+\theta}{n}\Big)+i\sin{\Big(\frac{2k\pi+\theta}{n}\Big)}}\Bigg);\text{ } k=0,1,...,n-1 \end{equation} $\\$ Now \begin{equation} \arg(z) = \tan^{-1}\Bigg(\tan\Bigg(\frac{2k\pi+\theta}{n}\Bigg)\Bigg) = \frac{2k\pi+\theta}{n} \end{equation}

Here we get values from $\frac{\theta}{n}$ to $ 2\pi-\frac{2\pi-\theta}{n}$ with an equal space of $\frac{2\pi}{n}$ i.e. $2\pi$ radians are divided into $n$ equal sections .

\begin{equation} \end{equation}

The solutions will be of the form \begin{equation} z = r^{1/n}e^{i(2k\pi+\theta)/n}; \text{ } k=0,1,...,n-1 \end{equation}

So, it is actually a geometric progression not an arithmetic progression.

Haotian H.
  • 41
Suzuna Minami
  • 262
  • Thanks for your answer John. However, I don't understand, if the argument of the solutions progress by $2k\pi/n$, isn't that an arithmetic sequence? – Haotian H. Jan 09 '20 at 06:02
  • 1
    Consider the below series: \begin{equation} a,aq,aq^2,...,aq^{n-1} \end{equation} Here, $a=(re^{i\theta})^{\frac{1}{n}}$, and $q=e^{\frac{2\pi}{n}}$ – Suzuna Minami Jan 09 '20 at 07:28
  • 1
    The phases are indeed in arithmetic progression, but solutions ain't. – Suzuna Minami Jan 09 '20 at 07:32
0

All solutions to $z^n=w$ must have the same magnitude, namely, $\sqrt[n]{w}$. Furthermore, recall that multiplying two complex numbers together multiplies their magnitudes and adds their arguments. In particular, raising a number to the $n$-th power multiplies its argument by $n$. This implies that any solution to $z^n=w$ must have an argument that, when multiplied by $n$, gives the argument of $w$, plus some multiple of $2\pi$. These values clearly form an arithmetic sequence, with difference $\frac{2\pi}n$. This explains your phenomenon.

ViHdzP
  • 4,582
  • 2
  • 18
  • 44