TITLE: Is exponential function analytic over all complex numbers
QUESTION [7 upvotes]: In my textbook, I find a text where it says $e^z$ is analytic everywhere (in complex plane). Is it true? If so, what is the proof? 
I approached using maclaurin series, which gives $e^z= \sum_{n=0}^{\infty} \frac{z^n}{n!}$. But given the Mac-laurin series, how can we find the points where it is analytic, i.e the points of convegrnece of its Taylor series? 
I am just a beginner here, so it my seem silly to some. However, any help will be appreciated. Thanks in advance! :D

REPLY [4 votes]: Theorem
Let $f = u + iv$ be defined on a domain $D \in C$. Then $f$ is analytic in $D$ if and only if $u(x, y)$ and $v(x, y)$ have continuous first partial derivatives on $D$ that satisfy the Cauchy-Riemann equations.

If $f(z) = e^z = e^{x+iy} = e^x \cos y + ie^x \sin y$. Then
$$u_x (x, y) = e^x \cos y = v_y(x, y),$$
$$v_x (x, y) = e^x\sin y = -v_x(x, y).$$
Thus the Cauchy-Riemann equations are satisfied, and in addition, the functions
$u_x , u_y , v_x , v_y$ are continuous in $C$. Therefore, the function $f$ is analytic in $C$, thus entire.