Showing $f(x,y)=(e^x \cos y,e^x \sin y)$ is $f:\mathbb{R}^2 \to \mathbb{R}^2 \setminus (0,0)$ Show $f(x,y)=(e^x \cos y,e^x \sin y)$ is $f:\mathbb{R}^2 \to  \mathbb{R}^2 \setminus  (0,0)$
So I need to show that for all $(b_1,b_2)$ there is $(a_1,a_2)$ such that
$b_1=e^{a_1} \cos a_2$   and  $b_2=e^{a_1} \sin a_2$?
How can I show it?
 A: Let $(b_1,b_2) \in \mathbb{R}^2 \setminus  (0,0)$.
Take log on $r$ in the polar coordinates.
$$r={\sqrt {b_1^{2}+b_2^{2}}}$$
$$a_2 =\operatorname {atan2} (b_2,b_1),$$
where atan2 is a common variation on the arctangent function defined as
$$
\operatorname {atan2} (y,x)={\begin{cases}\arctan({\frac {y}{x}})&{\mbox{if }}x>0\\\arctan({\frac {y}{x}})+\pi &{\mbox{if }}x<0{\mbox{ and }}y\geq 0\\\arctan({\frac {y}{x}})-\pi &{\mbox{if }}x<0{\mbox{ and }}y<0\\{\frac {\pi }{2}}&{\mbox{if }}x=0{\mbox{ and }}y>0\\-{\frac {\pi }{2}}&{\mbox{if }}x=0{\mbox{ and }}y<0\\{\text{undefined}}&{\mbox{if }}x=0{\mbox{ and }}y=0\end{cases}}$$
Then $b_1 = r \cos(a_2)$ and $b_2 = r \sin(a_2)$.  Take $a_1 = \ln r$ so that $b_1 = e^{a_1} \cos(a_2)$ and $b_2 = e^{a_1} \sin(a_2)$.  (Note that $r > 0$ since $(b_1,b_2) \ne (0,0)$)
A: You have to show that $\forall (a,b) \ne (0,0)$ we can find $(x,y)$ such that:
$$
a=e^x \cos y \quad \mbox{and} \quad b=e^x\sin y
$$ 
dividing the two equation we find:
$$ \tan (y)=\frac{b}{a}$$
That gives a value for $y$ for all $a \ne 0$
and squaring and adding the two equation we find:
$$a^2+b^2=e^{2x}$$
that has a solution $x=\frac{1}{2}\ln(a^2+b^2)$ for all $a,b \ne 0$
The case $a=0$ and $b > 0$ gives $ y=\frac{\pi}{2}+2k\pi$ and $x=\ln b$.
The case $a=0$ and $b < 0$ gives $ y=-\frac{\pi}{2}+2k\pi$ and $x=\ln |b|$.
