Finding complex roots of $(z-i)^n-(z+i)^n =0$

Question

I have a problem. First part of it was to find $z$ such that $$\left|\frac{z-i}{z+i}\right| = 1$$ and I've quickly figured out, that $z \in \Bbb R$.

How ever, now I have to (using previous part of the problem) find all roots of the following equation: $$(z-i)^n - (z+i)^n = 0$$

I have no idea how to do this. I tried adding the second part to the right side, then dividing by it, but with no success.

Answer 1

Rearrange as follows:

$$(z-i)^n=(z+i)^n \\ \left(\frac{z-i}{z+i}\right)^n=1$$

Let $w=\frac{z-i}{z+i}$. The equation $w^n=1$ is easy to solve, but you also know that since $|w|=1$, i.e. $\left|\frac{z-i}{z+i}\right|=1$, $z \in \mathbb R$.

Answer 2

Hint: $\qquad (z-i)^n - (z+i)^n=0\iff \Bigl(\dfrac{z-i}{z+i}\Bigr)^n=1.$

Can you solbe for $\;u=\dfrac{z-i}{z+i}$?

Answer 3

Observe that $(z-i)^n=(z+i)^n$ gives $\left|(z-i)^n\right|=\left|(z+i)^n\right|$, and so, as for all $z$, $|z|^n=\left|z^n\right|$, $|z-i|=|z+i|$;

so as you have proven previously, $z$ must be real.

Observe then that $$\arg\left((z+i)^n\right)=\arg\left((z-i)^n\right);$$ but as $\arg(z+i)=-\arg(z-i)$, we find that $\arg\left((z+i)^n\right)=n\times \arg(z+1)$ is either $0$ or $\pi$ radians.

Can you solve it from here?

Answer 4

$$(z-1)^n-(z+1)^n=0\ (n\ \epsilon\ \mathbb{Z}^+)$$ $$\Rightarrow (z-1)^n=(z+1)^n \Rightarrow \left(\frac{z-1}{z+1}\right)^n=1=1\angle 2\pi k\ (k=0,1,2,...n-1)$$ $$\Rightarrow \frac{z-1}{z+1}=1\angle \frac{2\pi k}{n}\Rightarrow 1 -\frac{2}{z+1}=1\angle \frac{2\pi k}{n}$$ $$\frac{2\pi k}{n}=\theta \Rightarrow 1 -\frac{2}{z+1}=1\angle \theta \Rightarrow z=\frac{1+1\angle \theta}{1-1\angle \theta}$$ $$\Rightarrow z = \frac{1+i\sin \theta + \cos \theta}{1-i\sin \theta - \cos \theta} = \frac{(1+i\sin \theta + \cos \theta)^2}{1-(i\sin \theta + \cos \theta)^2}$$ $$\Rightarrow z= \frac{2\cos^2 \theta + 2i\sin \theta \cos \theta +2i\sin \theta + 2\cos \theta}{2\sin^2 \theta -2i\sin \theta \cos \theta}$$ $$\Rightarrow z=\frac{2(i\sin \theta + \cos \theta)(\cos \theta + 1)}{-2i\sin \theta (i\sin \theta + \cos \theta)}=\frac{i(\cos \theta +1)}{\sin \theta}$$ $$\Rightarrow z = \csc \theta + i\cot \theta$$ $$\Rightarrow z = \csc \frac{2\pi k}{n} + i\cot \frac{2\pi k}{n}\ (k=0,1,2,...n-1)$$