I have some trouble with an algebraic excercise about unitary mappings. 

**First part**

If $f: \mathbb{R} \rightarrow \mathbb{R} $ is unitary, how to proof that $f$ can only be the identity?

*My attempts so far*

A unitary mapping $f:\mathbb{Z} \rightarrow \mathbb{Z}$ is the identity, because
$$
f(n) = f(1+1+ \cdots +1) = f(1)+f(1)+ \cdots +f(1)=1+1+\cdots+1=n
$$
This means that a unitary mapping $f: \mathbb{Q} \rightarrow \mathbb{Q}$ is the identity, as shown below.
$$f(q)=f(ab^{-1})=f(a)f(b)^{-1}=ab^{-1}=q$$
I didn't know how to deal with $\mathbb{R}$, but I thought:
$$x = x-\lfloor x \cdot 10^n \rfloor \cdot 10^{-n} + \lfloor x \cdot 10^n \rfloor \cdot 10^{-n}$$
The rightmost term is rational, so we can write
$$f(x)=f(x-\lfloor x \cdot 10^n \rfloor \cdot 10^{-n}) + \lfloor x \cdot 10^n \rfloor \cdot 10^{-n} $$
We can act the same way by ceiling x.
$$x = \lceil x \cdot 10^n \rceil \cdot 10^{-n} - (\lceil x \cdot 10^n \rceil \cdot 10^{-n}-x)$$

The leftmost term is rational, so $f(x) = \lceil x \cdot 10^n \rceil \cdot 10^{-n} - f(\lceil x \cdot 10^n \rceil \cdot 10^{-n}-x)$. 

For large $n$ the expressions $\lceil x \cdot 10^n \rceil \cdot 10^{-n}-x$ and $x-\lfloor x \cdot 10^n \rfloor \cdot 10^{-n}$ will be small, but I don't know if $f(\lceil x \cdot 10^n \rceil \cdot 10^{-n}-x)$ and $f(x-\lfloor x \cdot 10^n \rfloor \cdot 10^{-n})$ behave the same way. Is this approach a good one or is there an easier one?

**Second part**

When this matter is proved, I can ask my second question: Is the only unitary mapping $f:\mathbb{C} \rightarrow \mathbb{C}$ the identity? 

This is how far I came. We know that $f(a+bi) =f(a)+f(b)f(i)= a+bf(i)$, so if $f(i)=i$, we are done. But is this true? Intuitively I say yes, because we the rotating and stretching up the complex plane are not unitary, and a projection on the real axis is not a multiplacitive mapping, and I see no other options. How should this be done?

I am looking forward to your answers guys.