Group homomorphism/R to C/e^it/Kernel and image/Example

We consider the analytic mapping

$\mathbb {R} \longrightarrow \mathbb {C} ,t\longmapsto e^{{\mathrm {i} }t}=\cos t+{\mathrm {i} }\sin t.$

Due to the exponential law (or the addition theorems for the trigonometric functions), we have ${}e^{{\mathrm {i} }(t+s)}=e^{{\mathrm {i} }t}e^{{\mathrm {i} }s}$. Therefore, this is a group homomorphism from the additive group ${}(\mathbb {R} ,+,0)$ into the multiplicative group ${}(\mathbb {C} ^{\times },\cdot ,1)$. We determine the kernel and the image of this mapping. To determine the kernel, we must identify those real numbers ${}t$ fulfilling

$\cos t=1\,\,{\text{ and }}\,\,\sin t=0.$

Because of the periodicity of the trigonometric functions, this is the case if and only if ${}t$ is an integer multiple of ${}2\pi$. Hence, the kernel is the subgroup ${}2\pi \mathbb {Z}$. For a point in the image, we have ${}\vert {e^{{\mathrm {i} }t}}\vert =\sin ^{2}t+\cos ^{2}t=1$; therefore, the image point belongs to the complex unit circle. The trigonometric functions run through the complete unit circle, so that the image group is the complex unit circle with its complex multiplication.