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## Represent the real numbers as complex numbers of modulus 1

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.12

Let $G$ be the additive group of real numbers and $H$ the multiplicative group of complex numbers with absolute value 1. Let $\varphi : G \rightarrow H$ be the homomorphism $r \mapsto e^{2 \pi i r}$. Find the kernel of $\varphi$ and the fibers over $-1$, $i$, and $e^{4 \pi i / 3}$.

Solution: By Euler’s formula we have $$e^{2 \pi i r} = \cos(2 \pi r) + i \sin(2 \pi r).$$ Thus for each integer $n$, we have $\varphi(r + n) = \varphi(r)$.

(1) If $\varphi(r) = 1$, we have $\sin(2 \pi r) = 0$. Thus $r$ is an integer or a half integer. But also $\cos(2 \pi r) = 1$, so that $r$ must be an integer. Thus $\mathsf{ker}\ \varphi = \mathbb{Z}$.

(2) If $\varphi(r) = -1$, we have $\sin(2 \pi r) = 0$. Thus $r$ is an integer or a half integer. But also $\cos(2 \pi r) = -1$, so that $r$ must be a half integer. Thus $\varphi^\star(-1) = \{ n + 1/2 \ |\ n \in \mathbb{Z} \}$.

(3) If $\varphi(r) = i$, we have $\cos(2 \pi r) = 0$. Thus $r$ is either a quarter integer or a three-quarter integer (i.e. $n + 1/4$ or $n + 3/4$ for some integer $n$.) Since $\sin(2 \pi r) = 1$, we see that $r$ must be a quarter integer. Thus $\varphi^\star(i) = \{ n + 1/4 \ |\ n \in \mathbb{Z} \}$.

(4) If $\varphi(r) = e^{4 \pi i / 3} = \cos(4 \pi / 3) + i \sin(4 \pi / 3)$, we have $\cos(2 \pi r) = \cos(4 \pi / 3)$. Thus we have $2 \pi n \pm 4 \pi / 3 = 2 \pi r$ for some integer $n$, and thus $r = n \pm 2/3$. Since $\sin$ is an odd function, the condition $\sin(4 \pi / 3) = \sin(2 \pi r)$ eliminates the negative solution. So $$\varphi^\star(e^{4 \pi / 3}) = \{ n + 2/3 \ |\ n \in \mathbb{Z} \}.$$