Let $\alpha=\phi+i\theta$. Simplify $|e^\alpha|$ and $|e^{i\alpha}|$. I know that the answers are $e^\phi$ and $e^{-\theta}$.
I think the first one can be solve as following:
$|e^\phi|=|exp(\phi)||cos(\theta)+i sin(\theta)||=|exp(\phi)|=e^\phi$.
(1) I'm not entirely convinced why $|cos(\theta)+i sin(\theta)|=1$. And (2) I'm not sure how to solve the second part. Any advice are appreciated.
Best Answer
I will first of all assume $\theta$ and $\phi$ are real numbers.
For real numbers $x,y\in \mathbb R,$ $|x+iy|$ is defined to be $\sqrt{x^2+y^2}.$
So, to answer your first question, $|\cos\theta+i\sin\theta|=\sqrt{\cos^2\theta+\sin^2\theta}=\sqrt1=1$ (Pythagorean theorem.)
Now, $|e^{i\alpha}|$ can be found the same way: $|e^{i\alpha}|=|e^{i(\phi+i\theta)}|=|e^{-\theta}||e^{i\phi}|=e^{-\theta}.$