the relation $f^2+g^2=1$ is followed only by trigonometric functions for differentiable functions

Question

Let $U\subseteq \mathbb{C}$ be an open, convex and connected subset and let $f,g:U\to \mathbb{C}$ be two differentiable functions so that $f',g'$ are continuous functions ($f,g \in C^1$).

I want to show that if $f^2+g^2=1$ over $U$ then there is a function $z:U\to \mathbb{C}$ ($z\in C^1$) such that $f(t)=\sin(z(t))$ and $g(t) = \cos(z(t))$.

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My try (the unimportant part):

I took this question from a book, in which it's said to use the following statement (which I've already proved):

if $f:U\to \mathbb{C}$ ($f\in C^1$) with $f(U)\neq \mathbb{C}$ then there exists a function $g:U\to \mathbb{C}$ ($g\in C^1$) and $C_0\in \mathbb{C}$ so that $f(t)=\exp(g(t))+C_0$.

For my question it was advised that I should use this statement for the functions $f+ig , f-ig$

I'm not sure how to continue, please help me.

Answer 1

Note that $f^2+g^2=1\iff(f+ig)(f-ig)=1$. In particular, $f+ig$ has no zeros. So, there's a function $h$ such that $f+ig=\exp\circ h$. Besides$$f-ig=\frac1{f+ig}=\frac1{\exp\circ h}=\exp\circ(-h).$$Therefore,$$f=\frac{f+ig+f-ig}2=\frac{\exp\circ h+\exp\circ(-h)}2$$and$$g=\frac{f+ig-(f-ig)}{2i}=\frac{\exp\circ h-\exp\circ(-h)}{2i}.$$So, take $z=-ih(\iff h=iz)$.

Answer 2

Given that $f^2+g^2=1$ you know that $$1=f^2+g^2=(f+gi)(f-gi).$$ In particular $f+gi\neq0$ and $f-gi\neq0$. Hence there exist functions $u$ and $v$ such that $$f+gi=\exp(u)+C_0\qquad\text{ and }\qquad f-gi=\exp(v)+C_1.$$ It follows that $$f=\frac{\exp(u)+\exp(v)+C_0+C_1}{2}\qquad\text{ and }\qquad g=\frac{\exp(u)-\exp(v)+C_0-C_1}{2i}.$$ Of course because $(f+gi)(f-gi)=1$ you know that $$1=(\exp(u)+C_0)(\exp(v)+C_1)=\exp(u+v)+C_0\exp(v)+C_1\exp(u)+C_0C_1,$$ which easily implies that $C_0=C_1=0$ and $v=-u$. So the above becomes $$f=\frac{\exp(u)+\exp(-u)}{2}=\sin(u)\qquad\text{ and }\qquad g=\frac{\exp(u)-\exp(-u)}{2i}=\cos(u).$$