$f(x,y)=(e^x \cos(y), e^x\sin(y))$ is one-to-one proof. And onto? Let $A:=\mathbb{R} \times (0,2\pi)$. Show that the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ defined by $f(x,y)=(e^x \cos(y), e^x\sin(y))$ is one-to-one on $A$. Is it onto?
My attempt
(one-to-one) Let $f(x_1,y_1)=f(x_2,y_2)$. Then
$$e^{x_1}\cos y_1=e^{x_2}\cos y_2$$ and
$$e^{x_1}\sin y_1=e^{x_2} \sin y_2$$
Squaring both and then adding we get:
$$e^{2x_1}\cos^2y_1+e^{2x_1}\sin^2 y_1=e^{2x_2}\cos^2y_2+e^{2x_2}\sin^2 y_2$$
$$=e^{2x_1}(\cos^2y_1+\sin^2y_1)=e^{2x_2}(\cos^2y_2+\sin^2y_2)$$
$$e^{2x_1}(1)=e^{2x_2}(1)$$
$$x_1=x_2$$
Not sure how to get $y_1=y_2$ here...Also not sure how to intuitively know if this is onto (and provide such proof). Thanks for the help! 
 A: You know that $x_1=x_2$, and you also know that $f(x_1,y_1)=f(x_2,y_2)$. So, this means that $f(x_1,y_1)=f(x_1,y_2)$. Then
$$e^{x_1}cos(y_1) = e^{x_1}cos(y_2)$$
and also
$$e^{x_1}sin(y_1) = e^{x_1}sin(y_2)$$
Since $e^{x_1}>0$ for all values of $x_1$ from the definition of the exponential function, multiplying both sides of each equations above by $\frac{1}{e^{x_1}}$ yields
$$\begin{cases}cos(y_1) = cos(y_2) \\ sin(y_1) = sin(y_2)\end{cases}$$
But since we are considering  $y\in ]0, 2\pi[$, if they have the same $cos$ and $sin$ simultaneously, then this implies that 
$$y_1 = y_2$$
So, $(x_1,y_1) = (x_2, y_2)$, then $f$ is one-to-one in $A$.
OBS: You can actually check the last statement by checking the quadrants in the unit circle. You know that:
$$\begin{cases}cos(y_1) = cos(y_2) \\ sin(y_1) = sin(y_2)\end{cases}$$
And also that both $y_1,y_2\in]0,2\pi[$.Then, by analyzing all the cases:
If $sin(y_1)=sin(y_2) \geq 0$, if:


*

*$cos(y_1)=cos(y_2) \geq 0$, then both $y_1,y_2\in]0,\frac{\pi}{2}]$, but both functions ($sin,cos$) are one-to-one in this interval, so this implies that $y_1=y_2$.

*$cos(y_1)=cos(y_2) < 0$, then both $y_1,y_2\in]\frac{\pi}{2},\pi]$, but both functions ($sin,cos$) are one-to-one in this interval, so this implies that $y_1=y_2$.
If $sin(y_1)=sin(y_2) < 0$, if:


*

*$cos(y_1)=cos(y_2) \geq 0$, then both $y_1,y_2\in[\frac{3\pi}{2},2\pi[$, but both functions ($sin,cos$) are one-to-one in this interval, so this implies that $y_1=y_2$.

*$cos(y_1)=cos(y_2) < 0$, then both $y_1,y_2\in]\pi,\frac{3\pi}{2}]$, but both functions ($sin,cos$) are one-to-one in this interval, so this implies that $y_1=y_2$.
