This story has a long personal history. In my first attempted complex analysis exam I was asked if the entire function has any fixed points. I had no idea how to approach this question and I failed the exam. This first failure was soon overcome but that particular question was —much to my advantage— never addressed during my study again. Much later I found several ways to solve this question.
The image of is invariant under a shift . The “Little Picard” theorem then asserts that this function maps onto since it cannot omit just a single value and so it must have a root. Another way uses “Great Picard” and the fact that only has a single root so it must attain the value infinitely often. Both approaches seem nice enough but depend on non-trivial theorems (“Great Picard” more so than “Little Picard”) and there is no indication of the location of the fixed points. A more pedestrian approach shows that there must be infinitely many fixed points along the curve but this is hardly in the spirit of complex analysis. This was the state of affairs until this week. Then I found a simple and much more satisfying answer.
The Banach fixed point theorem asserts that every contraction of a complete metric space has a single fixed point. I will use that theorem in the following setting.
Let be a non-empty closed convex subset of an open set and let be holomorphic. If there exists a constant such that on then restricted to is a contraction and therefore has a single fixed point in .
For let denote the horizontal strip
and let be the branch of the logarithm that maps the slit complex plane onto the interior of . If then
on so the conditions of the theorem above apply and therefore has a single fixed point (which must lie in the interior.) Each is a fixed point of . Unfortunately the same argument does not work for where has two more fixed points.
The fixed point method works equally well for a number of other functions. Here are two more examples:
Example 1. Let and . The branch of that maps the (closed) upper half plane onto the half strip
is a contraction on . So has exactly one fixed point in each such strip (and therefore also exactly one in its complex conjugate).
Example 2. Let and odd. The branch of that maps the complement of the open unit disc into the strip
is a contraction on . So has exactly one fixed point in each such strip.
These example for , and work so well because their inverses have a nice derivative that is less than one except in a small bounded region. Put in another way these functions all satisfy a differential equation of the form for some and some polynomial .