# Complex equation proof of existence of roots

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 $$z$$).

• I changed your notation to the standard $\bar z$ for conjugation. – Clayton Feb 2 at 20:44

\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.

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.

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.