How to find the closed form of a complex number in polar form?

Question

To find the third roots of $z=5-\sqrt{2}i$ I determined the polar form of $z$ obtaining $\sqrt{27}\text{cis}(344.20\overline{7})$.

This yielded the three roots $w_1=\sqrt{3}\text{cis}(114.73\overline{6})$, $w_2=\sqrt{3}\text{cis}(234.73\overline{6})$, $w_3=\sqrt{3}\text{cis} (354.73\overline{6})$.

In normal form this is $w_1=−0.7247448715 + 1.573132185i$, $w_2=−1 − 1.414213562i$ and $w_3=1.724744872 − 0.15891862i$. Recognizing $w_2=-1-\sqrt{2}i$ permits to find also the closed forms of $w_3$ and $w_1$ by multiplying $w_2$ with $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)^2$ respectively.

How does one find the closed forms of the roots if one doesn't recognize one of them?

In the case of a third root it is possible to express $\cos(\alpha/3)$ with $\cos(\alpha)$ by solving a cubic equation. But to find e.g. that $1-2i$ is a fifth root of $41+38i$ this doesn't work anymore.

Answer 1

If you are asking on how to solve them without trig, then you could do it quite simply as follows:

$$5-\sqrt2i=(a+bi)^3=a^3+3a^2bi-3ab^2-b^3i$$

Thus,

$$5=a^3-3ab^2$$

$$\sqrt2=b^3-3a^2b$$

Which is a system of equalities which result in a cubic, which upon solving, gives

$$w_1=\frac{1-\sqrt6}2+\frac{(\sqrt6-1)(3\sqrt6+8)}{10\sqrt2}i\\\vphantom{\cfrac11}w_2=-1-\sqrt2i\\w_3=\frac{1+\sqrt6}2+\frac{(\sqrt6+1)(3\sqrt6-8)}{10\sqrt2}i$$

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In general, if you are trying to find all $n$th roots of $x+yi$, just set

$$x+yi=(a+bi)^n$$

and solve from there.

Answer 2

A symbolic solution, Im not sure if this is useful as an answer: if we normalize the number $z$ to $w:=z/|z|$, then we have that $w=\frac5{27}-i\frac{\sqrt 2}{27}=\cos\alpha+i\sin\alpha$, because $|z|=27$ and $|w|=1$, hence

$$z^{1/3}=\left(27 e^{i(\alpha+2\pi\Bbb Z)}\right)^{1/3}=\sqrt[3]{27}e^{i\frac{\alpha+2\pi\Bbb Z}{3}}$$

with $\alpha=\arctan\left(\frac{\sin \alpha}{\cos \alpha}\right)=\arctan\frac{-\sqrt2}{5}$.

By the other side: $z$ is a $n$-th root of $z^n$.

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EDIT: ok, I think that I understand what are you asking. Let some complex number $z:=a+ib$, then one $n$-th root of $z$ is $w_1=c+id$ such that $d/c=b/a$ (same angle) and $(a^2+b^2)^n=c^2+d^2$ ($n$-th root of the norm).

Once you knows $w_1$ the set of $n$-th roots of $z$ is

$$z^{1/n}=\left\{w_k=w_1\left(\cos\frac{2\pi}n (k-1)+i\sin\frac{2\pi }n(k-1)\right):k\in\Bbb Z\land \frac{2\pi}n(k-1)\in(-\pi,\pi]\right\}$$

Of course it remains the problem to found and easy expression for the sine and cosine of angles of the form $2\pi k/n$ (Im not sure if this can be solved easily considering the transcendental nature of trigonometric functions, I mean that probably these quantities are not necessarily algebraic numbers). I hope it helps.