Is exponential function analytic over all complex numbers 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  
 A: The radius of convergence of the power series
$$ \sum\limits_{n=0}^{\infty}a_{n}z^{n} $$
is given by the ratio test to be
$$ R = \frac{1}{\limsup |a_{n}|^{1/n}} $$
In your case, the radius of convergence is $\infty$ and hence the series defining the exponential converges for all complex numbers and as the power series is analytic in its disk of convergence, you have that the exponential is analytic on the entire complex plane.
To see that radius of convergence is indeed $\infty$ in your case, observe that $\lim\limits_{n\to \infty} (n!)^{\frac{1}{n}} = \infty$ and $a_{n} = (n!)^{-1}$ in your case. A proof of this limit can be found elsewhere on this forum.
A: 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.
