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Given a complex number $z$, how can I solve $$(z+i)^n+(z-i)^n=0$$ for z?

My attempt

I thought that I couldn't possibly convert this into Euler's format. So, I took to Binomial Theorem. $$(z+i)^n+(z-i)^n=0$$ Using binomial theorem, $$2\left[{n \choose 0}z^n+{n \choose 2}z^{n-2}i^2+{n \choose 4}z^{n-4}i^4+...\right]=0$$ $${n \choose 2}z^{n-2}+{n\choose 6}z^{n-6}+...={n \choose 0}z^{n}+{n \choose 4}z^{n-4}+...$$ How can I solve this after this step?

Any hint is appreciated.

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    Rewrite as $\frac{z + i}{z - i} = (-1)^{\frac{1}{n}}$. So the question essentially boils down to finding the $n^{th}$ roots of -1. Think you can do that? – Anon Mar 30 '21 at 03:56
  • https://math.stackexchange.com/questions/607487/the-roots-of-the-equation-zn-1zn OR https://math.stackexchange.com/questions/1562603/solve-the-equation-z13iz-13-0 – lab bhattacharjee Mar 30 '21 at 03:57
  • @labbhattacharjee Not the same. That has nth roots of 1, this has nth roots of $-1$. – Anon Mar 30 '21 at 03:58
  • $n^{th}$ roots of $-1$ are $cis (\frac{(2k+1)\pi}{n} )$ for $k = 0, 1, ... n - 1$. – Anon Mar 30 '21 at 04:06
  • @Kaind, I've given two examples , one has $1$ and the other $-i$ – lab bhattacharjee Mar 30 '21 at 04:08
  • Geometrically, $|z+i|=|z-i|$ implies $z$ is real. Substituting $z+i=re^{i\theta}$ yields $\cos(n\theta)=0$ (and also $r=\csc\theta$ by drawing a right triangle). – anon Mar 30 '21 at 04:47

1 Answers1

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Solving:

$$\begin{align}y&=\frac{z+i}{z-i }\\ yz-yi&=z+i \\ (y-1)z&=(y+1)i \\ z&=\frac{iy+i}{y-1} \end{align}$$

Then you need $y^n=-1.$

Thomas Andrews
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