Solve complex equation $\left(1+z\right)^{2n}+\left(1-z\right)^{2n}=0$

Question

How do you solve this complex equation?

$$\left(1+z\right)^{2n}+\left(1-z\right)^{2n}=0$$

Answer 1

As $1-z\ne0$

$$\left(\dfrac{1+z}{1-z}\right)^{2n}=-1=e^{(2m+1)\pi i}$$

$(1+z)/(1-z)=e^{(2m+1)\pi i/2n}$ where $0\le m\le2n-1$

Apply https://www.qc.edu.hk/math/Junior%20Secondary/Componendo%20et%20Dividendo.htm

Then divide the numerator and the denominator by $e^{(2m+1)\pi i/4n}$

Finally use Intuition behind euler's formula

Answer 2

Since $(1+z)^{2n}=-(1-z)^{2n}$, $|1+z|^{2n}=|1-z|^{2n}$ and $$|1+z|=|1-z|.$$ If $\Re(z)=a$, this further implies $(1+a)^2=(1-a)^2$ and $$a=0.$$ Therefore, $\overline{z}=-z$, and $(1+z)^{2n}+\overline{(1+z)^{2n}}=0$, which is equivalent to $$\Re\left((1+z)^{2n}\right)=0,$$ or, $$2n\arg(1+z)\equiv\frac{\pi}{2}\pmod{\pi}.$$ The solutions to $z$ will therefore be $$z\in\left\{i\tan\left(\frac{k\pi}{4n}\right)\,\middle|\, -(2n-1)\leq k\leq 2n-1,\, k\text{ odd}\right\}.$$

Answer 3

Rearrange the equation,

$$\dfrac{1-z}{1+z}=(-1)^\frac{1}{2n}=e^{i a}$$

where $a=\frac{2k+1}{2n}\pi$, $k$ = 0,1, ... 2n-1. Then, solve for $z$,

$$z=\frac{1-e^{ia}}{1+e^{ia}}= -i\tan\frac a2$$