# Complex Exponential is Uniformly Continuous on Half-Planes

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## Theorem

Let $a\in\R$.

Then $\exp$ is uniformly continuous on the half-plane $\set {z \in \C : \map \Re z \le a}$.

### Corollary

Let $X$ be a set.

Let $\family {g_n}$ 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 $\map \Re {\map g x} \le a$ for all $x \in X$.

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

## Proof

Let $\epsilon>0$.

For $x, y \in \C$ with $\map \Re x, \map \Re y \le a$:

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

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

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

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

Thus if $\cmod {x - y} < \delta$, $\cmod {e^x - e^y} < e^a \epsilon$.

Thus $\exp$ is uniformly continuous.

$\blacksquare$