Show $(\mathbb{C}^*,\cdot,1) \cong (\mathbf{T},\cdot,1)\times (\mathbb{R}_{>0},\cdot,1)$ where $\mathbf{T} = \{a+bi \in \mathbb{C}|a^2+b^2=1\}$. Show $(\mathbb{C}^*,\cdot,1) \cong (\mathbf{T},\cdot,1)\times (\mathbb{R}_{>0},\cdot,1)$ where $\mathbf{T} = \{a+bi \in \mathbb{C}|a^2+b^2=1\}$ and $\mathbb{C}^*=\mathbb{C}\setminus \{0\}$.
So this means we need to show that the complex plane without the origin is isomorph to a cylinder of radius 1 'just above' the complex plane. Now if we can find a bijection $f:\mathbb{C}^* \to\mathbf{T}\times \mathbb{R}_{>0}$ where $f(x\cdot y) = f(x)\cdot f(y)$ we are done. I was thinking of using the function $$f:(a+bi) \mapsto (e^{i\tan^{-1}(\frac{b}{a})},a^2+b^2)$$ which simply means that we map every point in $\mathbb{C}^*$ to a point in the cylinder with angle equal to the angle of the original point and height equal to the magnitude of the original point. I got this idea from thinking about cylindrical coordinates $(r,\theta , z)$ where $r=1$. This is certainly a bijection because there exist an inverse transformation $f^{-1}: (u,r) \mapsto \left(r \left(\cos(\frac{\log(u)}{i}\right) + i \sin\left(\frac{\log(u)}{i}\right)\right)$ leading back to the original point in $\mathbb{C}^*$. However if we let $x=a+bi$ and $y=c+di$:
$$f(x\cdot y) = (e^{i\tan^{-1}(\frac{ad+bc}{ac-bd})}, (ac-bd)^2+(ad+bc)^2)$$
and
$$f(x)\cdot f(y)= (e^{i\tan^{-1}(\frac{b}{a})},a^2+b^2) \cdot (e^{i\tan^{-1}(\frac{d}{c})},c^2+d^2)$$
$$ = (e^{i[\tan^{-1}(\frac{b}{a})+\tan^{-1}(\frac{d}{c})]},(a^2+b^2)\cdot (c^2+d^2))$$
How do I prove from here that $f(x\cdot y) =f(x)\cdot f(y)$? I have a hunch that's it is actually true and would be a lot easier to prove if converted to polar coordinates but I am trying to avoid this (should I?). The statement is true if $\tan^{-1}(\frac{ad+bc}{ac-bd} )= \tan^{-1}(\frac{b}{a}) +  \tan^{-1}(\frac{c}{d})$. Thanks in advance for any help, and I hope the question is clear!
 A: Are you confident with direct product of groups?
If so, consider the following result (which can be easily proved): given a group $G$, if there exist normal subgroups $H,K$ of $G$ such that $H\cap K=\{1\}$ and $G=HK\overset{def}{=}\{hk\in G: h\in H,\ k\in K\}$, then $G\cong H\times K$.
Now set $G:=(\mathbb{C}^{*},\ \centerdot,\ 1)$, $H:=(\mathbf{T},\ \cdot,\ 1)$ and $K:= (\mathbb{R}_{>0},\ \cdot,\ 1)$. Since $G$ is abelian, $H$ and $K$ are normal subgroups of $G$ and trivially $\mathbf{T}\cap \mathbb{R}_{>0}=\{1\}$. Moreover, due to the polar representation of complex numbers (see the "Complex numbers" section here if needed and notice that for each $\theta\in \mathbb{R},\ \vert e^{i\theta}\vert =1$), we also have $HK=G$ and then we are done.
Here is a proof of the mentioned result. So we have a group $G$ and normal subgroups $H,K$ with $H\cap K=\{1\}$ and $HK=G$. This means that for each $g\in G,\ \exists h\in H$ and $\exists k\in K$ such that $hk=g$. Moreover if $h_{1}\in H$ and $k_{1}\in K$ are such that $hk=h_{1}k_{1}$, then $h_{1}^{-1}h=k_{1}^{-1}k$ and therefore $h_{1}^{-1}h=k_{1}^{-1}k\in H\cap K\iff h_{1}^{-1}h=k_{1}^{-1}k=1$. Then $h_{1}^{-1}h=1\implies h_{1}^{-1}=h^{-1}\implies h_{1}=h$ and similarly we get $k_{1}=k$. This argument says exactly that the map $$
\alpha\colon H\times K\longrightarrow G=HK,\qquad (h,k)\mapsto hk$$ is a bijection. (By the way, notice that this $\alpha$ is precisely the map Hagen von Eitzen suggested). Now we need to show that $\alpha$ is also a homomorphism of groups, i.e. that for each $(h,k),(h_{1},k_{1})\in H\times K$ one has $$hh_{1}kk_{1}=\alpha ((h,k)(h_{1},k_{1})=\alpha((h,k))\alpha((h_{1},k_{1}))=hkh_{1}k_{1}$$
To prove this equality we use normality of $H,K$ and the fact that their intersection is trivial. Indeed,
$$
hh_{1}kk_{1}=hkh_{1}k_{1}\iff hh_{1}kk_{1}k_{1}^{-1}h_{1}^{-1}k^{-1}h^{-1}=1\iff hh_{1}kh_{1}^{-1}k^{-1}h^{-1}=1.
$$
Now, since $K$ is normal in $G$, $h_{1}kh_{1}^{-1}\in K$ (a subgroup $K$ of a group $G$ is normal in $G$ iff $\forall k\in K,\ \forall g\in G$ one has $gkg^{-1}\in K$) and hence $hh_{1}kh_{1}^{-1}k^{-1}h^{-1}=h(h_{1}kh_{1}^{-1}k^{-1})h^{-1}\in K$ ($K$ is a subgroup). In the same way, normality of $H$ in $G$ implies that $kh_{1}^{-1}k^{-1}$ belongs to $H$ and then also $l:=hh_{1}kh_{1}^{-1}k^{-1}h^{-1}=hh_{1}(kh_{1}^{-1}k^{-1})h^{-1}$ does. Therefore, $l\in H\cap K$ and then $l=1$, so that $\alpha$ actually is a morphism of groups.
A: Try $\mathbf T\times \mathbb R_{>0}\to\mathbb C^\times $, $(a+bi, r)\mapsto r(a+bi)$
