magnitude of complex exponental always equals 1?

Question

as we all know $$e^{j\theta} = \cos\theta + j\sin\theta \\ |e^{j\theta}| = \sqrt{\cos^2\theta + \sin^2\theta} = 1$$

That means $|e^{j\theta}| = 1$ with any value $\theta$ is ($2\pi, \frac{\pi}{3}$, etc). However, my teacher said $$e^{-j\pi k} = (e^{-j\pi})^k$$ and he said that $(e^{-j\pi}) = -1$. I really don't understand why this is. Can anybody explain this for me.

Answer 1

Algebraically, substituting $x = \pm\pi$ into Euler's Formula, $e^{ix} = \cos x + i \sin x$ yields:

$$\begin{align}e^{\pm i\pi} &= \cos(\pm\pi) + i \sin (\pm\pi) \\ &= -1 + 0\cdot i\\ &= -1\end{align}$$

Geometrically, $e^{\pm i\pi}$ is an anti-clockwise/clockwise rotation of the point corresponding to $1$ on the Argand diagram by $\pi$ radians, which results in the point corresponding to $-1$.

To answer your confusion about why $e^{i\pi} = -1$ even though $|e^{i\theta}| = 1$, do note that $e^{i\pi} \not = |e^{i\pi}|$. It should be $|e^{i\pi}| = |-1| = 1$.

Answer 2

It won't always be equal to one. Consider the following case: $$ re^{j\theta}=r(\cos\theta+j \sin\theta). $$