Solving a Sum of Complex Exponentials

Question

Suppose I have the following equation.

$$e^x+e^{\omega x}+e^{\omega^2 x}=0$$

where $\omega=e^{2i\pi/3}$. How do I find all solutions to this equation in the complex plane? Do I need to use numerical techniques or are there algebraic ways to solve this? I think the answer to the algebraic is "no" but I was wondering if there was a tried and true method.

Also, what about the variants below

$$e^x+\omega^2 e^{\omega x}+\omega e^{\omega^2 x}=0$$

$$e^x+\omega e^{\omega x}+\omega^2 e^{\omega^2 x}=0$$

Certainly, $x=0$ is a solution to these, but how would i find others?

Answer 1

The roots are invariant under multiplication by $\omega$. Here are implicit plots of the real (in blue) and imaginary (in red) parts of $e^x+e^{\omega x} + e^{\omega^2 x}$. The roots are where the red and blue curves intersect: evidently those are points (fairly regularly spaced) along the negative real axis and the rays at angles $\pm 2\pi/3$ from that.

The first few negative roots are approximately $$-1.84981279919014, -5.44123335502365, -9.06899753487163, -12.6965955465468, -16.3241942781214, -19.9517930065763$$

Answer 2

your first sum is equal to $${{\rm e}^{x}}+2\,{{\rm e}^{-x/2}}\cos \left( 1/2\,x\sqrt {3} \right) =0$$ i think a numerical method will help here one solution is given by $$x\approx -1.849812799$$ your second equation is given by $${{\rm e}^{x}}-{{\rm e}^{-x/2}}\cos \left( 1/2\,x\sqrt {3} \right) - \sqrt {3}{{\rm e}^{-x/2}}\sin \left( 1/2\,x\sqrt {3} \right) =0$$ one solution is given by $$x\approx -7.859792867$$