Why are the roots for a complex number placed symmetrically around Argand diagram? Aren't we finding revolutions?

Question

I am currently learning about how to use De Moivre's Theorem to find complex roots. My textbook has stated that "polar form only gets you one possible angle or argument. To get all possible roots, you have to consider equivalent expressions to z which could involve going around one or more revolutions first. Therefore in general, $z^n = r \operatorname{cis} (\theta + 2k \pi)$, with the $2k\pi$ indicating $k$ revolutions".

So I'm guessing this means that the solutions are found by going once around the circle (full revolution$ - 2\pi$) over and over again.

However, in an example, the roots are not in the same place (as I would expect if they were revolutions); they are placed symmetrically around the Argand diagram.

But shouldn't they be in the same place, if we are doing revolutions over and over again?

Answer 1

This is best explained with congruences: the argument of a complex number is not a real number, but its congruence class modulo $2\pi$.

Thus, for instance, if you have to solve (as your figure seems to indicate) $$z^4=r\operatorname{cis}(\theta)$$ and if you denote $\rho$ the modulus and $\varphi$ the argument of $z$, De Moivre's formula yields the equalities $$\rho^4=r,\qquad 4\varphi\equiv \theta\pmod{2\pi}$$ whence, instantly, $$\rho=r^{1/4},\qquad \varphi\equiv \frac{\theta}4 \pmod{\frac{2\pi}4},$$ which means (modulo $2\pi$) that $$\varphi=\frac\theta 4+ \frac{k\pi}2 \quad (k=0\,1,2,3).$$ As you can see, the corresponding images in the Argand-Cauchy plane deduce from each other by rotations of angle $\pi/2$.

Answer 2

If you want the $nth$ root of $re^{i\theta}$, the roots are $r^\frac{1}{n}e^{i\frac{\theta+2k\pi}{n}}$, for $k=0,1,...,n-1$. You see that you keep rotating and then divide the angle by $n$ until you get n roots.

Answer 3

$$z^n = r \operatorname{cis} \theta = r \operatorname{cis} (\theta + 2km \pi)$$ Thus $$z = r^{1/n} \operatorname{cis} (\frac{\theta}{n} +m\frac{ 2k \pi}{n})$$The roots are separated from each other by an angle of $\frac{ 2k \pi}{n}$. The formula for $z$ is valid for any integer $m$ but it is enough to take $0 \le m \le n-1$ since using other integers for $m$ would only repeat values of $z$ already found for $m$ in the range $[0,n-1].$ The diagram you provided is for $r=4,n=4, \theta=\pi.$ It is correct and shows the 4 fourth roots of $-4i=4 \text {cis}\pi,viz$ $$\sqrt 2 \text { cis} \frac{\pi}{4},\sqrt 2 \text { cis} \frac{3\pi}{4},\sqrt 2 \text { cis} \frac{5\pi}{4},\sqrt 2 \text { cis} \frac{7\pi}{4}.$$