For a complex number $z = x + iy$, why does $\mid e^z \mid = e^x$? For a complex number $z = x + iy$, why does $\mid e^z \mid = e^x$ ?
 A: This is quite easy to be understood if you use Euler's formula
$$e^{x+iy} = e^xe^{iy} = e^x (\cos y + i \sin y)$$
so that $|e^z|= |e^x||e^{iy}| = |e^x|\cdot 1 = |e^x|$.
A: Assuming that the exponential of a complex number shares properties of the real exponential,
$$e^z=e^{x+iy}=e^xe^{iy}.$$
Now, using Taylor's development and evaluating the powers of $i$
$$e^{iy}=1+iy-\frac{y^2}2-i\frac{y^3}{3!}+\frac{y^4}{4!}+i\frac{y^5}{5!}-\frac{y^6}{6!}\cdots$$
The real and imaginary parts are respectively
$$\Re(e^{iy})=1-\frac{y^2}2+\frac{y^4}{4!}-\frac{y^6}{6!}\cdots$$
$$\Im(e^{iy})=y-\frac{y^3}{3!}+\frac{y^5}{5!}\cdots$$
where you can recognize Taylor's development of the functions $\cos(y)$ and $\sin(y)$. This explains why Euler's formula holds
$$e^{iy}=\cos(y)+i\sin(y).$$
Now
$$|e^z|=|e^x||e^{iy}|=e^x\sqrt{\cos^2(y)+\sin^2(y)}=e^x.$$

It is possible, though a little tedious, to show directly that 
$$\Re^2(e^{iy})+\Im^2(e^{iy})=1$$ without knowing what these functions are.
A: Recall that $|x+iy| = \sqrt{x^2+y^2}$ and $e^{iy} = \cos y + i\sin y$. Hence
$$|e^{x+iy}| = |e^x|\cdot|e^{iy}| = e^x\cdot|\cos y + i \sin y| = e^x\cdot\sqrt{\cos^2 y+ \sin^2 y} = e^x\cdot \sqrt{1} = e^x. $$
