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Spring 2009 - Problem 8

Picard's little theorem

Let $f\p{z}$ be an entire non-constant function that satisfies the functional equation

f\p{1 - z} = 1 - f\p{z}

for all $z \in \C$ . Show that $f\p{\C} = \C$ .

Solution.

Suppose there exists $w \notin f\p{\C}$ . This implies that $f\p{z} = w$ has no solutions. By a change of variables, this means $f\p{1 - z} = w$ has no solutions and so

w = f\p{1 - z} = 1 - f\p{z} \implies f\p{z} = 1 - w

has no solutions either. Thus, if $w \neq 1 - w$ , this implies that $f$ misses two points, which contradicts Picard's little theorem. Hence, it must be that $w = 1 - w \implies w = \frac{1}{2}$ , but this is also impossible as

f\p{1 - \frac{1}{2}} = 1 - f\p{\frac{1}{2}} \implies f\p{\frac{1}{2}} = \frac{1}{2}.

Thus, no $w$ could have existed to begin with, i.e., $f\p{\C} = \C$ , as required.