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Spring 2015 - Problem 12

entire functions

Find all entire functions f ⁣:C→C\func{f}{\C}{\C} that obey

f′(z)2+f(z)2=1.f'\p{z}^2 + f\p{z}^2 = 1.

Prove that your list is exhaustive.
Solution.

Notice that

(f′(z)+if(z))(f′(z)−if(z))=1,\p{f'\p{z} + if\p{z}}\p{f'\p{z} - if\p{z}} = 1,

so neither factors may vanish. Thus, f′+ff' + f admits a logarithm, i.e., there exists an entire function gg such that

eg(z)=f′(z)+if(z).e^{g\p{z}} = f'\p{z} + if\p{z}.

We also get

eg(z)(f′(z)−if(z))=1  ⟹  f′(z)−if(z)=e−g(z).e^{g\p{z}}\p{f'\p{z} - if\p{z}} = 1 \implies f'\p{z} - if\p{z} = e^{-g\p{z}}.

If h(z)=g(z)ih\p{z} = \frac{g\p{z}}{i}, then g(z)=ih(z)g\p{z} = ih\p{z}, so eih(z)=f′(z)+if(z)e^{ih\p{z}} = f'\p{z} + if\p{z}. This implies

f(z)=sin⁡h(z)andf′(z)=cos⁡h(z).f\p{z} = \sin{h\p{z}} \quad\text{and}\quad f'\p{z} = \cos{h\p{z}}.

Taking derivatives,

f′(z)=h′(z)cos⁡h(z)=h′(z)f′(z)  ⟹  f′(z)(1−h′(z))=0.f'\p{z} = h'\p{z}\cos{h\p{z}} = h'\p{z}f'\p{z} \implies f'\p{z}\p{1 - h'\p{z}} = 0.

If ff is constant, then f(z)2=1f\p{z}^2 = 1, so f(z)=1f\p{z} = 1 or f(z)=−1f\p{z} = -1, identically. Otherwise, we see that f′(z)f'\p{z} is zero only on an isolated set, which means that h′(z)=1h'\p{z} = 1 except possibly on an isolated set. But by continuity, it follows that h′(z)=1h'\p{z} = 1 everywhere, so h(z)=z+ch\p{z} = z + c for some constant cc.

In summary, if ff is constant, then f(z)=1f\p{z} = 1 or f(z)=−1f\p{z} = -1. Otherwise, f(z)=cos⁡(z+c)f\p{z} = \cos\p{z + c} for some constant cc, which completes the proof.