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I have a polynomial $p(z) = (z+i)^{100} - (z-i)^{100}$ and I am trying to show that all roots should be real. I was thinking I should use the Fundamental Theorem of Algebra which says that every non-constant polynomial has at least one zero in C.

I thought maybe I needed to prove that $p(z)$ has no complex coefficients and hence no zeros would be in C. However I am not sure if I am on the right track and if I am, where can I go from there. If the exponents were smaller, I would probably expand the function and check the coefficients like that. The number being too big here, I am guessing I am missing a more elegant solution.

Another idea would be to try to find a root or two and and see if there is a pattern whereby I can show that all roots are real. Is this possible? How would I proceed to find any such root?

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$p(z) = 0$ implies that $|z-i|^{100} = |z+i|^{100}$ and that is equivalent to $$ |z-i|^2 = |z+i|^2 \\ \iff (z-i)(\overline {z - i}) = (z+i)(\overline {z + i}) \\ \iff 2i(z - \overline z) = 0 $$ so that $z = \overline z$, i.e. $z$ is a real number.

Geometrically, $|z-i| = |z+i|$ is the locus of all points in the complex plane which have the same distance from $i$ and $-i$, and that is the real line.

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Let be more general and denote $p_n(x)= (z+i)^n-(z-i)^n$.

If $z$ is a root of $p_n$, we have $$\frac{z+i}{z-i}=\phi_k$$ where $e^{\frac{i2k\pi}{n}}=\phi_k$ is a $n$-th unit root. Therefore

$$z=i\frac{1+\phi_k}{1-\phi_k}=-\frac{1}{\tan\left(\frac{2k\pi}{n}\right)}$$

Proving the desired result.

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    $\begingroup$ For the last step — showing that $i \frac{1+\varphi_k}{1-\varphi_k}$ is real — a more geometric method is to diagram the points $0$, $1$, $1+\varphi_k$, $1-\varphi_k$ in the complex plane, and the circle of radius $1$ with center $1$. The angle between ${1+\varphi_k}$ and ${1-\varphi_k}$ is then the angle of an inscribed triangle on a diameter of a circle, so it’s a right angle. $\endgroup$ – Peter LeFanu Lumsdaine Oct 21 at 14:03
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(This answer was found based on the answer by mathcounterexamples.net, but simplifies the argument, and I hope makes it a bit more intuitive.)

The equation $(z+i)^n - (z-i)^n=0$ can be rewritten as $(z+i)^n = (z-i)^n$. Looking just at magnitude, this implies that $|z+i| = |z-i|$.

Looking at this geometrically in the complex plane, it’s now quite intuitive that $z$ must be real: if $z$ were in the upper half-plane, then $z+i$ would have greater magnitude than $z-i$, or vice versa if $z$ were in the lower half-plane.

To prove this intuition algebraically, write $z= a +ib$, with $a,b$ real; then expanding out the equation $|z+i|^2 = |z-i|^2$ in terms of $a$ and $b$ gives directly that $b=0$.

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  • $\begingroup$ @leftaroundabout Thanks for the typo catch! $\endgroup$ – Peter LeFanu Lumsdaine Oct 21 at 23:23

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