Show that $f(x,y) = (e^x\cos y, e^x \sin y)$ is one to one on $E := \{(x,y):x \in \mathbb{R} ,0 < y < \frac{\pi}{2} \}$.

Question

Show that $f(x,y) = (e^x\cos y, e^x \sin y)$ is one to one on $E := \{(x,y):x \in \mathbb{R} ,0 < y < \frac{\pi}{2} \}$.

It is relatively obvious, but I don't know how to write it down, like "there's no restriction in this interval?".

This question had been ask many times; however, all of them show why the function is not 1-1, but not why it is 1-1 on this specific interval.

Answer 1

Suppose $(x,y), (x',y') \in E$ and $f(x,y)=f(x',y')$. Take norm on both sides to get $e^{2x}=e^{2x'}$ which gives $x=x'$. Now we have $\cos y =\cos y'$ and $\sin y=\sin y'$. This implies that $y-y'$ is a multiple of $2\pi$. But since $y,y' \ (0,\pi /2)$ the only possibility is $y=y'$.