Prove that $\phi$ is surjective homomorphism.

Question

Let $G=(\mathbb{R},+)$ and $S^\prime=\{z \in \mathbb{C} \space| \space|z|=1\}$. Define $\phi : G\longrightarrow S^\prime$ by $\phi(x)=e^{2\pi ix}$

(a) Prove that $\phi$ is surjective homomorphism.

• $\phi$ is well define,

  $x=y$ ; $x,y \in \mathbb{R}$ $\implies \phi(x)=e^{2\pi ix}= e^{2\pi iy}=\phi(y) $

• $\phi$ is homomorphism,

  $\phi(x+y)= e^{2\pi i(x+y)}= e^{2\pi ix}\times e^{2\pi iy} =\phi(x)\phi(y)$

• $\phi$ is surjective,

  Let any $z \in S^{\prime}$ then $|z|=1$ and there exist $x \in \mathbb{R}$ s.t $z= \cos(2\pi x)+i\sin(2\pi x)$

  This follows $z=\phi(x)$

$\therefore$ $\phi$ is surjective homomorphism.

Here what I tried, Surjective part is not sure. Can anyone verify my answer?

Answer 1

You claimed that there is some $x$ such that $\phi(x)=z$, but you did not prove it.

Since $|z|=1$, if you write $z$ as $a+bi$, with $a,b\in\Bbb R$, then $a^2+b^2=1$. So, $b\in[-1,1]$. Let $\alpha=\frac{\arccos(a)}{2\pi}$. With this choice of $\alpha$, $\cos(2\pi\alpha)=a$. And\begin{align}b^2&=1-a^2\\&=1-\cos^2(2\pi\alpha)\\&=\sin^2(2\pi\alpha),\end{align}and therefore $\sin(2\pi\alpha)=\pm b$. If $\sin(2\pi\alpha)=b$, take $x=\alpha$; otherwise, take $x=-\alpha$.