Are the complex conjugates roots?

Question

Suppose that $s=a+bi$ is a complex root of $f(s) = 6^s+2^s-3^s$. Is the complex conjugate also a root? I have plotted the function and computed some numerical roots, and it seems that this is indeed the case, but can not find a proof for it. Any ideas? For instance $s=-0.296247533230636 + 1.91583650734156*I$ is a numerical root.

Answer 1

Use the fact that:

Proposition: $\forall z=re^{i\theta}\in \mathbb{C}$, $\overline{z}=re^{-i\theta}$

Proof: Write $z=r(\cos(\theta) + i\sin(\theta))$ and use the properties $\cos$ and $\sin$.

Then, notice how, if $s$ is a root, $\overline{f(s)}=0=e^{\overline{s}\ln(6)}+e^{\overline{s}\ln(2)}-e^{\overline{s}\ln(3)}=6^\overline{s}+2^\overline{s}-3^\overline{s}=f(\overline{s})$.