Let $ g,f: \mathbb C\longrightarrow\mathbb C$ be two continuous functions such that for each $ z\neq 0$, $ g(z)=f(\frac1z)$. Prove that there is a $ z\in\mathbb C$ such that $ f(\frac1z)=f(-\bar z)$
Yes, you can easily combine both maps as a map $F:S^2\rightarrow \mathbb C$ and from Borsuk-Ulam Theorem for such map there is $x\in S^2$ s.t. $F(x)=F(-x)$. Projecting $x$ from $S^2$ to $\mathbb C$ we reach to $ f(\frac{1}z)=f(-\bar z) $