Show that $f(\{(x,y):|y|<\pi/2\})=\{(u,v):u>0\}$

Question

Given the function $f(x,y)=(e^x\cos(y),e^x\sin(y))$, in the open set $\{(x,y):|y|<\pi/2\}$, $f$ has an inverse, since if $$u=e^x\cos(y),v=e^x\sin(y)$$ then $$x=\ln(u^2+v^2)^{1/2},y=\arctan(u/v)$$

How can I show that $f(\{(x,y):|y|<\pi/2\})=\{(u,v):u>0\}$?

Answer 1

It is clear that LHS $\subseteq$ RHS since $\cos t$ is positive in $(-\pi /2, \pi /2)$ and $e^{x} >0$. For the other inclusion let $u>0$ and define $x =\ln \sqrt{u^{2}+v^{2}}, y=\arctan (\frac v u)$. Check that $f(x,y)=(u,v)$. By definition of $\arctan$ we have $|y| <\frac {\pi} 2$.