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Solve the equations

(1) \( \displaystyle (z-i)^{n} - (z+i)^{n} = 0 \)

(2) \( \displaystyle (z-i)^{n} + (z+i)^{n} = 0 \)

where \( \displaystyle n\in\mathbb{N} \) and \( \displaystyle i \) is the imaginary unit.

Hello Nickos.

\(\displaystyle{1)}\) Let \(\displaystyle{z\in\mathbb{C}}\) such that \(\displaystyle{\left(z-i\right)^n-\left(z+i\right)^n=0}\) and \(\displaystyle{n\geq 2}\) .

For \(\displaystyle{n=1}\) the equation \(\displaystyle{\left(z-i\right)^n-\left(z+i\right)^n=0}\) has no solutions on \(\displaystyle{\mathbb{C}}\) .

Obviously, \(\displaystyle{z\neq \pm i}\) and then \(\displaystyle{\left(\dfrac{z+i}{z-i}\right)^n=1}\) , which means that

the complex number \(\displaystyle{w=\dfrac{z+i}{z-i}}\) is a solution of the equation \(\displaystyle{x^n=1\,,x\in\mathbb{C}}\) .

So, there is \(\displaystyle{k\in\left\{1,...,n-1\right\}}\) such that \(\displaystyle{\dfrac{z+i}{z-i}=e^{k}}\) , where

\(\displaystyle{e=\cos\,\dfrac{2\,\pi}{n}+i\,\sin\,\dfrac{2\,\pi}{n}}\) . Futhermore,

\(\displaystyle{\begin{aligned} \left|w\right|^2=1&\implies \left|z+i\right|^2=\left|z-i\right|^2\\&\implies \left|z\right|^2+1-i\,z+i\,\bar{z}=\left|z\right|^2+1-i\,zi\,\bar{z}\\&\implies 2\,i\,\left(z-\bar{z}\right)=0\\&\implies z=\bar{z}\\&\implies z\in\mathbb{R}\end{aligned}}\)

and \(\displaystyle{z=\dfrac{i+i\,e^{k}}{e^{k}-1}\,\,\,,e^{k}=\cos\,\dfrac{2\,k\,\pi}{n}+i\,\sin\,\dfrac{2\,k\,\pi}{n}\,,k\in\left\{1,...,n-1\right\}}\) .

I would like to make a question : Is the above solution complete ;

Let's see what happens for \(\displaystyle{n\in\left\{2,3\right\}}\) .

\(\displaystyle{\bullet\,n=2: \left(\dfrac{z+i}{z-i}\right)^2=e^2=\left(\cos\,\pi+i\,\sin\,\pi\right)^2=1\,,z=\dfrac{i+i\,e}{e-1}=0}\)

\(\displaystyle{\bullet\,n=3: z_1=\dfrac{i+i\,e}{e-1}=\dfrac{1}{\sqrt{3}}\,,z_2=-\dfrac{1}{\sqrt{3}}}\)

\(\displaystyle{2)}\) For each \(\displaystyle{z\in\mathbb{C}}\) holds :

\(\displaystyle{\left(z+i\right)^n+\left(z-i\right)^n=0\iff \left(\dfrac{z+i}{z-i}\right)^{n}=-1}\) .