Why is $\mathbb{R}/\mathbb{Z}$ isomorphic to the complex numbers of module one? Wikipedia states that the quotient group $\mathbb{R}/\mathbb{Z}$ is isomorphic to all complex numbers of module $1$. I have a hard time making sense out of this, and in particular, how complex numbers even come into play. What I am asking for, if possible, is an intuitive answer. Giving a bijective homomorphism is not considered a sufficient answer unless an intuitive explanation proves difficult.
 A: Because the homomorphism
$$
\varphi\colon\mathbb{R}\to \mathbb{C}\setminus\{0\}
$$
(the domain is a group with respect to addition, the codomain with respect to multiplication) defined by
$$
\varphi(t)=\cos(2\pi t)+i\sin(2\pi t)
$$
has the unit circle as its image and $\mathbb{Z}$ as its kernel.
A: $\mathbb R/\mathbb Z$ identifies each point on the real line with a unique point in the interval $[0,1)$. We can stretch this interval by a factor of $2\pi$ and wrap it around the circumference of the unit circle in the complex plane.
Because the real number $1$ maps to $0$ in the interval $[0,1)$ we find that joining the ends together works, and each real number can be identified with a number of circuits of the unit circle.
It is like treating the whole real line as a piece of elastic and wrapping it around the unit circle so that all the integers end up in the same place.
A: Hint :  $x \rightarrow e^{ix}$
A: In $\mathbb{R}/\mathbb{Z}$ numbers are equivalent if their difference
is in $\mathbb{Z}$.
For example: $1.2\equiv5.2$ and $\pi\equiv\pi+1$.
Every real number is equivalent to a number in $[0,1)$ ,since if
$a\in\mathbb{R}$ we can write $a$ as a decimal number$a=a_{0}.a_{1}a_{2}...$
and $a\equiv0.a_{1}a_{2}...\in[0,1)$.
Note that $0\equiv1$. This looks like a circle - We start at the
point $0$ and we continue on until we reach $1$ - which is equivalent
to $0$, thus closing the loop.
Now, if we draw the set of complex numbers with module $1$ in the
complex plain we also get a circle - this is because a circle is described
by $x^{2}+y^{2}=1$ and if we denote $z=x+iy$ we get that 
$$
|z|=1\iff\sqrt{x^{2}+y^{2}}=1\iff x^{2}+y^{2}=1
$$
A: To make it really intuitive: When dividing ourt $\mathbb Z$, you identifiy $0$ with $1$ 8and $2$ and $-42$ and ..., as well as identify $0.1$ with $1.1$ adn ...). If you want to physically do this identification with the real number line, intuitively represented by a thread of infinite length, you have to roll it back into itself like a coil and - by identification - produce a circle. The same unti circle we find in the ocmplex plane. However, this intuitive explanation may be intuitive, but I don't really think it is an  explanation (which is given by the properties of the exponential function, i.e. any solution of $f'(z)=f(z)$ with $f(0)=1$ is necessarily a group homomorphism from addition to multiplication, and is periodic).
A: Note that $\mathbb{R}/\mathbb{Z}$ is equal to the to the unit interval $[0,1]$, where the equivalence class of zero and one are equal. This can be interpreted as glueing the $0$ and $1$ together, in which case you get a circle. Now try to find the homomorphism, by using GA316's hint.
