Extending the exponential to complex

Question

We have the exponential function is defined from $\mathbb{R}$ to $\mathbb{R}$. However if we wanted to prove the existence of a continuous function $\exp:\mathbb{C}\rightarrow\mathbb{C}$ that is a function that extends $e^x$ to a complex domain and range, how would we go about it?

I know intuitively that: for $z\in\mathbb{C}$, letting in general $z=a+bi$ with $a,b\in\mathbb{Z}$ it holds that $e^z\in\mathbb{C}$. I also know that the exponential function is expanded with the general taylor series as $e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$ but this is strictly defined for the reals.

I am not sure how to go about this proof, it would seem that we would have to start with definining an entirely new function since the expansion only is defined on the reals.

Answer 1

For showing continuity of a function , it is enough to show that $lim_{z\to z_0} f(z) = f(z_0) $. Since , $e^z$ is defined on all over $\mathbb{C} $ , so, for each $z_0$, $lim_{z\to z_0} e^z $ exists and equal to $e^{z_0}$

Edit : as you say , you begin with the assumption , exponential defined on whole $\mathbb{R} $. Then, simply write $f(z)=e^z=e^x cos(y) + i (e^x sin(y)) = u(x,y)+i v(x,y) $. Clearly, as, $e^x,cos(y),sin(y) $ defined on whole $\mathbb{R} $, from this we can conclude that $u(x,y),v(x,y)$ are defined on $\mathbb{R}^2 $ , so, $f(z)$ defined on $\mathbb{C} $

Answer 2

Consider, for $z \in \mathbb C$, the power series

$$\sum_{k=0}^{\infty}\frac{z^k}{k!}.$$

It is easy to see that this power series has radius of convergence $= \infty$, hence it defines a function $f: \mathbb C \to \mathbb C$,

$$f(z)=\sum_{k=0}^{\infty}\frac{z^k}{k!}.$$

From the theory of power series it is known that $f$ is continuous on $ \mathbb C.$ Furthermore we have

$$f(x)=e^x$$

for all $x \in \mathbb R.$