4

Find all the entire functions (holomorphic functions on $\mathbb{C}$) $f,g$ satisfying $f^2+g^2=1$.

Of course $f,g$ can be constants satisfying $f^2+g^2=1$. But if they are not constants, do they exist?

Maybe Liouville Theorem can be used here to solve this problem, but how to use?

José Carlos Santos
  • 427,504
YZ Zhang
  • 145
  • 1
    For any entire function $h$ we can take $f(z)=\sin h(z),g(z)=\cos h(z)$. I suspect those might be all the examples. If we show $f$ is $\sin$ of some entire function $h$, it will follow $g$ is $\pm\cos h$, so replacing $h$ by $\pi-h$ if necessary we will have $g=\cos h$. – Wojowu Apr 10 '20 at 11:12
  • @Wojowu Is $\sin h$ holomorphic? (Sorry I haven't learnt about $\sin z$ defined in $\mathbb{C}$) And do there exsist other functions in other forms? – YZ Zhang Apr 10 '20 at 11:16
  • 1
    $\sin z$ can be defined by a convergent power series, so it is entire. Composing it with $h$, which is also entire, gives an entire function. – Wojowu Apr 10 '20 at 11:17
  • @Wojowu I got it, thank you! – YZ Zhang Apr 10 '20 at 11:18

1 Answers1

16

Since $f^2+g^2=1$, $(f+ig)(f-ig)=1$. So, $f+ig$ is an entire function without zeros, and therefore it can be written as $e^{hi}$, for some entire function $h$. So,$$f-ig=\frac1{f+ig}=\frac1{e^{hi}}=e^{-hi}.$$Therefore,$$f=\frac{f+ig+f-ig}2=\frac{e^{hi}+e^{-hi}}2=\cos\circ h$$and$$g=\frac{f+ig-(f-ig)}{2i}=\frac{e^{hi}-e^{-hi}}{2i}=\sin\circ h.$$

José Carlos Santos
  • 427,504