Motivation: Continuous functions $f: \mathbb{R}\to\mathbb{R}$ that are periodic have a fixed point.$^1$ This does not extend to the plane as $F: \mathbb{R}^2 \to \mathbb{R}^2$ defined by the rule, $$ F(x,y) = (\cos(x),y+1)$$ is a real-analytic function that is periodic and has no fixed points. One can restrict to entire functions to guarantee a fixed point. That is, if $f: \mathbb{C} \to \mathbb{C}$ is entire and periodic, then $f$ has a fixed point.$^2$
What are some different or weaker conditions than holomorphicity one can give that guarantees a fixed point for functions from the plane to the plane?
Proof of claim 1: Fix $f: \mathbb{R}\to\mathbb{R}$ continuous with period $c\in \mathbb{R}\setminus \left\{0\right\}$. Consider $g: \mathbb{R}\to \mathbb{R}$ defined by the rule $$g(x) = f(x) - x.$$ We see that there exists $n,k\in \mathbb{Z}$ distinct such that $$g(nc) = f(nc) - nc = f(0) - nc < 0 \text{ and } g(kc) = f(kc) - kc = f(0) - kc > 0.$$ By the intermediate value theorem, $f$ has a fixed point.
Proof of claim 2: Fix $f: \mathbb{C}\to\mathbb{C}$ entire with period $c\in \mathbb{C}\setminus\left\{0\right\}$. Suppose for contradiction that there does not exist a fixed point for $f$. It follows that $f(z) \neq z + c$ for any $z$. Consider $g: \mathbb{C}\to \mathbb{C}$ defined by the rule $$g(z) = f(z) - z.$$ Clearly $g$ is entire and $g\left(z\right) \neq 0$ for any $z$. Moreover, $g \left(z\right) \neq c$ for any $z$. By Picard's little theorem, we have that $g$ is constant. This implies $f$ is not periodic, which is a contradiction.