the relation f2+g2=1 is followed only by trigonometric functions for differentiable functions Let U⊆C be an open, convex and connected subset and let f,g:U→C be two differentiable functions so that f,g are continuous functions f(,g∈C1) I want to show that if f2+g2=1 over U then there is a function z:U→C such that f(t)=sin⁡(z(t)) and g(t)=cos⁡(z(t))

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the relation f2+g2=1 is followed only by trigonometric functions for differentiable functions   Let U⊆C be an open, convex and connected subset and let f,g:U→C be two differentiable functions so that f,g are continuous functions f(,g∈C1)  I want to show that if f2+g2=1 over U then there is a function z:U→C such that f(t)=sin⁡(z(t))  and  g(t)=cos⁡(z(t))

Mazzuranavf · Accepted Answer

Note that f2+g2=1⇔(f+ig)(f−ig)=1. In particular, f+ig has no zeros. So, there&#039;s a function h such that f+ig=exp⋅h. Besidesf−ig=1f+ig=1exp∘h=exp∘(−h)Therefore,f=f+ig+f−ig2=exp∘h+exp∘(−h)2andg=f+ig−(f−ig)2i=exp∘h−exp∘(−h)2iSo, take z=−ih(⇔h=iz)

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

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