How does the complex exponential function transform the unit circle?

Question

I know you can write every complex number on the unit circle as $e^{i\theta} = \cos(\theta)+i\sin(\theta).$ But what does it look like when you raise $e$ to the values? You get $e^{\cos(\theta)+i\sin(\theta)} = e^{\cos(\theta)}(\cos(\sin(\theta)) + i\sin(\sin(\theta))),$ but I'm having trouble visualizing this.

Answer 1

$e^{\cos\theta+i\sin\theta}$ can be explained as follows.

i) $e^{i\theta}$ is a rotation about $\theta$ radians in the complex plane. $e^{\cos\theta+i\sin\theta}=\exp({e^{i\theta}})$ is the a rotation in $\mathbb{C}^1$ represented in the space $X$, where $\exp:\mathbb{C}^1\to X$.

ii) The range of $\cos\theta$, and $\sin\theta$ is $[-1,1]$. So, as you've expanded $e^{\cos\theta+i\sin\theta}$ as $$e^{\cos(\theta)+i\sin(\theta)} = e^{\cos(\theta)}(\cos(\sin(\theta)) + i\sin(\sin(\theta)))$$ $e^{\cos\theta+i\sin\theta}$ could also mean a rotation in $\mathbb{C^1}$ by $\cos\theta$, multiplied a re-scaling factor of $(\cos(\sin(\theta)) + i\sin(\sin(\theta)))$. So for example, when acted upon the vector $\vec{x}$, $\exp({e^{i\theta}})\vec{x}$ means a rotation by $\cos\theta$ of the new vector $(\cos(\sin(\theta)) + i\sin(\sin(\theta)))\vec x$.

I may not have covered all the possible interpretations, but I hope this is useful.