Prove that a map is Injective

Question

How can I prove that $f: (0, \infty) \times (0,\pi) \to \mathbb{R}^2$ where $f(x,y) = (\sinh(x)\sin(y),\cosh(x)\cos(y))$ is injective?

Answer 1

If $f$ is not injective, then there exist $a\not=c$, $b\not=d$ such that

$\cosh(a) \cos(b)=\cosh(c) \cos(d)\\ \sinh(a) \sin(b)=\sinh(c) \sin(d)$

W.L.O.G let $a>c$. Then $\sinh(a)>\sinh(c)$ and $\cosh(a)>\cosh(c)$ since both functions are increasing in $(0,\infty)$. Therefore $\sin(b)<\sin(d)$ and $\cos(b)<\cos(d)$, which can only occur in $(0,\pi)$ if $\cos(b)<0$ and $\cos(d)>0$ (see graph), but then the above could not be true because $\cosh(a)$ is positive.

Answer 2

Suppose $\cosh a\cos b=\cosh c\cos d$ and $\sinh a\sin b=\sinh c\sin d$. Square the two relations: \begin{gather} \cosh^2 a\cos^2 b=\cosh^2 c\cos^2 d\\ \sinh^2 a\sin^2 b=\sinh^2 c\sin^2 d \end{gather} Sum and subtract the two: \begin{gather} \cosh^2 a + \sinh^2 a = \cosh^2 c + \sinh^2 c\\ \cos^2 b - \sin^2 b = \cos^2 d - \sin^2 d \end{gather} Therefore $$ \cosh 2a = \cosh 2c $$ which means that $2a=2c$ because $\cosh$ is injective in $(0,\infty)$.

Furthermore $$ \cos 2b = \cos 2d $$ that is, $2b=2d$ or $2b=2\pi-2d$, from which $$ b=d\quad\text{or}\quad b=\pi-d $$ The second possibility, together with $\cosh a\cos b=\cosh c\cos d$, gives $\cos b=\cos(\pi-b)$, that is $\cos b=0$ and, again, $b=d=\pi/2$.

Answer 3

Let $u:=\sinh x\sin y$, $\>v:=\cosh x\cos y$, $\>z:=x+iy$. Then $$u-iv=-i\sinh x\sinh(iy)-i\cosh x\cosh(iy)=-i\cosh(x+iy)=-i\cosh z\ .$$ We have to prove that $\cosh z_1\ne\cosh z_2$ when $z_1\ne z_2$ and both $z_1$ and $z_2$ lie in the strip $S:\ 0<{\rm Im}(z)<\pi$.

Put $e^{z_i}=:w_i\ne0$. Then $\cosh z_1=\cosh z_2$ is equivalent with $$(w_1-w_2)\left(1-{1\over w_1 w_2}\right)=0\ .$$ Since $\exp$ is injective on $S$ the assumption $z_1\ne z_2$ implies $w_1\ne w_2$. Furthermore, $w_1w_2=1$ would imply $z_1+z_2=2k\pi i$ for some $k\in{\mathbb Z}$. But there are no $\>z_1$, $z_2\in S$ satisfying this equation. Altogether it follows that $\cosh z_1=\cosh z_2$ is impossible for two different $z_i\in S$.