# Complex Exponential is Uniformly Continuous on Half-Planes

## Theorem

Let $a\in\R$.

Then $\exp$ is uniformly continuous on the half-plane $\{z\in\C : \Re(z) \leq a\}$.

### Corollary

Let $X$ be a set.

Let $(g_n)$ be a family of mappings $g_n : X\to\C$.

Let $g_n$ converge uniformly to $g:X\to\C$.

Let there be a constant $a\in\R$ such that $\Re( g(x)) \leq a$ for all $x\in X$.

Then $\exp g_n$ converges uniformly to $\exp g$.

## Proof

Let $\epsilon>0$.

For $x,y\in\C$ with $\Re (x),\Re (y)\leq a$,

\(\displaystyle \vert e^x-e^y\vert\) | \(=\) | \(\displaystyle \vert e^y\vert\cdot \vert e^{x-y}-1\vert\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e^{\Re (y)}\cdot \vert e^{x-y}-1\vert\) | Absolute Value of Complex Exponential | ||||||||||

\(\displaystyle \) | \(\leq\) | \(\displaystyle e^{a}\cdot \vert e^{x-y}-1\vert\) | Exponential is Strictly Increasing |

Because Exponential Function is Continuous, there exists $\delta>0$ such that $\vert e^{z}-1\vert<\epsilon$ for $|z|<\delta$.

Thus if $|x-y|<\delta$, $\vert e^x-e^y\vert < e^a\epsilon$.

Thus $\exp$ is uniformly continuous.

$\blacksquare$