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If $f, g: \mathbb{C} \to \mathbb{C}$ are analytic functions that satisfy $(f(z))^{2} + (g(z))^{2} = 1$ for all $z \in \mathbb{C}$, show that exist an analytic function $h: \mathbb{C} \to \mathbb{C}$ such that $f = cos(h)$ and $g = sin(h)$.

Someone can help me? Thank you in advance!

J.A.G
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1 Answers1

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Following @r9m's hint, write $f+ig=e^{ih}$ so $f-ig=\frac{1}{f+ig}=e^{-ih}$. Then $f=\frac{e^{ih}+e^{-ih}}{2}=\cos h$; similarly, $g=\sin h$. Note that since $f,\,g$ aren't real-valued, $f\pm ig$ aren't in general conjugate pairs, so $h$ isn't real in general.

J.G.
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  • How can I see that exist a function $h$ such that $f + ig = e^{ih}$? – J.A.G Oct 21 '19 at 19:45
  • @JaimeGrimalAlves Well, $f+ig\ne0$, so you just need to address subtleties in the definition of $\ln(f+ig)$, then multiply by $-i$. – J.G. Oct 21 '19 at 20:04