Simplify $|e^\alpha|$ and $|e^{i\alpha}|$.

Question

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.

Answer 1

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: $|e^{i\alpha}|=|e^{i(\phi+i\theta)}|=|e^{-\theta}||e^{i\phi}|=e^{-\theta}.$

Answer 2

In general, given a complex number $z$, $|e^z|= e^{Re(z)}$. Indeed, if we put $z=a+ib$:

$$|e^{z}|=|e^{a}e^{ib}|=|e^a||e^{ib}|= |e^a|$$

The reason why $|\cos(\alpha) + i \sin(\alpha)|=1$ is that you can consider the complex number $w=\cos(\alpha) + i \sin(\alpha)$ and use the fact that $|w|= \sqrt{w \overline{w}}$.

Answer 3

With

$\alpha = \phi + i \theta, \tag 1$

we have

$\vert e^\alpha \vert = \vert e^{\phi + i \theta} \vert = \vert e^\phi \vert \vert e^{i\theta} \vert$ $= \vert e^\phi \vert \vert \cos \theta + i \sin \theta \vert = \vert e^\phi \vert \sqrt{\cos^2 \theta + \sin^2 \theta} = \vert e^\phi \vert = e^\phi, \tag 2$

since

$0 < e^\phi \in \Bbb R; \tag 3$

also,

$i\alpha = i\phi + i(i\theta) = -\theta + i \phi, \tag 4$

so that

$\vert e^{i\alpha} \vert = \vert e^{-\theta + i\phi} \vert = \vert e^{-\theta} e^{i\phi} \vert = \vert e^{-\theta} \vert \vert \cos \phi + i \sin \phi \vert; \tag 5$

at this point the calculation follows (2) from $\vert e^\phi \vert \vert \cos \theta + i \sin \theta \vert$ closely and we conclude that

$\vert e^{i \alpha} \vert = e^{-\theta}. \tag 6$

Note that we have used the standard definition

$\forall u, v \in \Bbb R, \; \vert u + iv \vert = \sqrt{u^2 + v^2}, \tag 7$

in evaluating (2), (6). This may in fact be viewed as a case of the Pythagorean theorem if the $u$- and $v$-axex are taken to be orthogonal to one another, and $\vert u + iv \vert$ is taken to be the hypoteneuse of the triangle formed by three points $(0, 0)$, $(u, 0)$ and $(u, v)$ in the $uv$ plane when considered as $\Bbb R^2$.