Why does $|e^{ix}| = 1$ when $x$ is a real number

Question

I'm halfway through a question on Complex Analysis and part of the solution says that $|e^{ix}|=1$ when $x$ is real.

But I cannot seem to find an explanation anywhere for this? I tried expanding it into trigonometric functions but got nowhere...

Answer 1

Recall that if $z = x + iy$ is a complex number, with $x$ and $y$ real, then $|z| = \sqrt{x^2 + y^2}$.

Recall also that $e^{ix} = \cos x + i \sin x$.

Therefore: $$ |e^{ix}| = |\cos x + i\sin x| = \sqrt{\cos^2x + \sin^2x} = \sqrt1 = 1$$

Answer 2

It's a typo. It should read $|e^{ix}|=1$ for all $x\in \Bbb{R}$.

Answer 3

The simple answer here is that $e^{ix}$ is on a circle of radius unity in the complex plane for any real $x$. Hence, its absolute value is unity.