Demonstration of Modulus of exp(z)

Question

I failed to find any step-by-step demonstration of the following equality:

$|e^z| = e^x$

Feedback: I was doing something really stupid: In the calculus of the module I was using $i^2$, resulting in $|e^x|\sqrt{\cos^2y - \sin^2y}$

Answer 1

With

$z = x + iy, \tag 1$

we have

$\vert e^z \vert = \vert e^{x + iy} \vert = \vert e^x e^{iy} \vert = \vert e^x \vert \vert e^{iy} \vert; \tag 2$

now,

$\vert e^{iy} \vert = \vert \cos y + i \sin y \vert = \sqrt{\cos^2 x + \sin^2 y} = \sqrt 1 = 1, \tag 3$

and, since $e^x > 0$ for all $x \in \Bbb R$,

$\vert e^x \vert = e^x; \tag 4$

assembling (3) and (4) into (2) yields

$\vert e^z \vert = e^x \tag 5$

as required.

Answer 2

Hint: If $z= x+ iy$ ($x$ and $y$ real), then $$e^z= (e^x\cos y) + i(e^x\sin y).$$ What is the magnitude of the right side?