Finding the distinct roots of a fractional complex number

Question

If I have to find the roots of the equation $$\Bigl(\frac{1-w}{1+w}\Bigr)^4 = 1, \text{where} \hspace{5pt} w \in \mathbb{C},$$ then I can simply let the term $$\frac{1-w}{1+w} = s. $$ Therefore $$s^4 = 1,$$ which means by inspection that $$s = i, -i, 1, -1.$$ But if I would like to use the formula $$w= re^{i\theta}$$ and/or $$w= r^{\frac{1}{n}}\biggl(\cos\biggl(\frac{\theta}{n} +\frac{2k \pi}{n}\biggr) + i \sin\biggl(\frac{\theta}{n} +\frac{2k \pi}{n}\biggr)\biggr)$$ to solve for these roots, how would I go about doing so? I know that $n = 4$, so for the four distinct roots, $k = 0, 1, 2 ,3$. After this, I don't know how to proceed.

Answer 1

It seems that you expect a benefit of expressing $w$ in polar form (and as a fourth root), but nothing substantiates this. On the opposite, I guess that it will completely obscure the resolution.

You can of course solve

$$s^4=1$$ in polar form, with

$$r^4e^{4i\theta}=1=e^{i2n\pi},$$

giving

$$s=e^{in\pi/2}$$ or $$s=1,i,-1,-i$$ but you already know this.

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A better option is to rewrite as

$$w^4-4w^3+6w^2-4w+1=w^4+4w^3+6w^2+4w+1$$ which simplifies as

$$w^3+w=0$$ or $w=0,w=\pm i$.

Answer 2

So you have $${1-w\over 1+w} = s$$ for $s\in \{1,-1,i,-i\}$. Now solving this on $w$ we get $$w={1-s\over 1+s}$$

So:

if $s=1$ you get $w=0$

if $s=-1$ then $w$ does not exists

if $s=i$ then $w = {(1-i)^2\over 2} = -i$

and

if $s=-i$ you get $w=i$.