Showing $f(x,y)=(e^x \cos y,e^x \sin y)$ is $f:\mathbb{R}^2 \to \mathbb{R}^2 \setminus (0,0)$

Question

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?

Answer 1

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 $(b_1,b_2) \ne (0,0)$)

Answer 2

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 $x=\ln |b|$.