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How can we try to prove that the equation $$ \left(\bar z+\frac{i}{z} \right)^{16} +\left(\bar z-\frac{i}{z} \right)^{16} =0 $$ has any solution (where $\bar z$ is a conjugate of $z$).

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    $\begingroup$ I changed your notation to the standard $\bar z$ for conjugation. $\endgroup$ – Clayton Feb 2 at 20:44
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$$\begin{aligned}\left(\bar z+\frac{i}{z} \right)^{16} +\left(\bar z-\frac{i}{z} \right)^{16} &= z^{-16}\left[\left(\bar z z+i \right)^{16}+\left(\bar z z-i \right)^{16}\right]\\&=z^{-16}\left[\left(|z|^2+i \right)^{16}+\left(|z|^2-i \right)^{16}\right]\\&=z^{-16}\left[\left(|z|^2+i \right)^{16}+\left(\overline{|z|^2+i }\right)^{16}\right]\end{aligned}$$ Set $w=|z|^2+i=re^{i\varphi}.$
Solve $$w^{16}+\overline{w}^{16}=0$$ or equivalently (because clearly $r\neq0$)$$e^{16i\varphi}+e^{-16i\varphi}=0$$ or $$\cos (16\varphi)=0,$$ which is easy to solve.

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If $\left(\bar z+\frac{i}{z} \right)^{16} +\left(\bar z-\frac{i}{z} \right)^{16} =0$ then we can write for $z \neq 0$ $\left(z \bar z+i \right)^{16} +\left(z \bar z-i \right)^{16} =0$

If we write $z = \rho e^{i \phi}$ then $z \bar z = \rho^2$ so the previous equation can be written $(\rho^2 + i)^{16} + (\rho^2 - i)^{16} = 0$

On the complex plane these are two vectors symmetric with respect to the real axis.

enter image description here

Two complex numbers like this have a 0 sum only when they are aligned with the imaginary axis. Moreover the 16th power are vectors rotating clockwise and counterclockise 16 times.

So the geometric condition for the $\alpha$ angle will be $\rho^2 \tan \alpha = 1$ and moreover $16 \alpha = \frac{\pi}{2}$ hence $\alpha = \frac{\pi}{32}$

So the magnitude must be $\rho = \sqrt{\frac{1}{\tan{\frac{\pi}{32}}}}$

In these conditions, the solution to the equation will be the set of complex numbers having magnitude $\rho$ computed as seen before, i.e. a circle around the origin of the complex plane.

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