Complex numbers involving roots of unity

Question

Let $z\in \mathbb{C}$ and $n\in \mathbb{N}, n \geq 1$. Solve the following equation: $$z+z^2+\dots+z^n=n|z|^n$$ Obviously, $(z,n)=(0,n)$ and $(1,n)$ are solutions, $\forall n \geq 1$. Considering $z \neq 0,1$ I tried to use $$z+z^2+\dots+z^n=z(1+z+\dots+z^{n-1})=z(z-\epsilon)(z-\epsilon^2)\dots(z-\epsilon^{n-1})$$ where $\epsilon=\cos \frac{2\pi}{n}+i\sin \frac{2\pi}{n}$, such that $$z(z-\epsilon)(z-\epsilon^2)\dots(z-\epsilon^{n-1})=n|z|^n$$ I then tried to take the norm on both sides and obtain some sort of inequalities, but didn't manage to get somewhere.

Answer 1

Expanding on the work of @WB-man by looking at the solutions graphically.

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$n=2$ $$ f(z) = z+ z^2 -2 \left| z\right|^{2} $$ Find the roots of the real and imaginary components. Solve $$ \begin{align} \text{Re } f &= x - x^2 - 3 y^2 = 0, \\ \text{Im } f &= y + 2 x y2 = 0, \\ \end{align} $$ The solution for the real component is plot in $\color{blue}{blue}$, imaginary component in $\color{red}{red}$.

The solutions are the intersection of the $\color{blue}{blue}$ and $\color{red}{red}$.

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$n=3$ $$ f(z) = z + z^2 + z^3 - 3 \left| z\right|^{3} $$ Solve $$ \begin{align} \text{Re } f &= -3 x^2 \sqrt{x^2+y^2}-3 y^2 \sqrt{x^2+y^2}+2 x^2+x-2 y^2 = 0, \\ \text{Im } f &= y + 4 x y = 0, \\ \end{align} $$

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$n=4$ $$ f(z) = z + z^2 + z^3 + z^{4} - 4 \left| z\right|^{4} $$ Solve $$ \begin{align} \text{Re } f &= x + 2 x^2 - 3 x^4 - 2 y^2 - 14 x^2 y^2 - 3 y^4 = 0, \\ \text{Im } f &= y + 4 x y + 4 x^3 y - 4 x y^3 = 0, \\ \end{align} $$

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$n=5$ $$ f(z) = z + z^2 + z^3 + z^{4} + z^{5} - 5 \left| z\right|^{5} $$ Solve $$ \begin{align} \text{Re } f &= x^7+x^6-21 x^5 y^2+x^5-15 x^4 y^2+x^4+35 x^3 y^4-10 x^3 y^2+15 x^2 y^4-6 x^2 y^2-7 y^6 \sqrt{x^2+y^2}-21 x^2 y^4 \sqrt{x^2+y^2}+2 x^2-7 x^6 \sqrt{x^2+y^2}-21 x^4 y^2 \sqrt{x^2+y^2}-7 x y^6+5 x y^4+x-y^6+y^4-2 y^2 = 0, \\ % \text{Im } f &= 5 x^4 y+4 x^3 y-10 x^2 y^3-4 x y^3+4 x y+y^5+y = 0, \\ \end{align} $$