morphism $df : Lie(\Bbb R, +) \to Lie(S^1)$

Question

Consider a morphism $f : (\Bbb R, +) \to \Bbb S^1$.

Prove $df : \Bbb R → Lie(\Bbb S^1 ) = i\Bbb R$ is of the form $df : t → 2πiαt, α ∈ \Bbb R$.

My attempt:

$df : Lie(\Bbb R)=\Bbb R → Lie(\Bbb S^1)= i\Bbb R$ is a linear map, therefore: $df(x)=df(x.1)=x.df(1)$ so $df$ is myltiplication by a scalar. But I don't know why it is of the form $2πiαt, α ∈ \Bbb R$?

Thank you for your help.

Answer 1

$df(1)\in i\mathbb{R}$ it is of the form $ic, c\in\mathbb{R}$ let $\alpha={c\over {2\pi}}$, $df(1)=2\pi i\alpha$ and $df(t)=df(1).t=2\pi i\alpha t$.