Complex Exponential Tends to Zero
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Theorem
Let $\exp z$ be the complex exponential function.
Then:
- $\ds \lim_{\map \Re z \mathop \to +\infty} e^{-z} = 0$
where $\map \Re z$ denotes the real part of $z$.
Proof
Let $z = x + iy$.
Let $\epsilon > 0$.
By the definition of limits at infinity, we need to show that there is some $M > 0$ such that:
- $x > M \implies \size {e^{-z} - 0} < \epsilon$
But:
\(\ds \size {e^{-z} - 0}\) | \(=\) | \(\ds \size {e^{-z} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \size {e^{-x} }\) | Modulus of Exponential is Modulus of Real Part | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {e^{-x} - 0}\) |
so we need an $M$ such that:
- $x > M \implies \size {e^{-x} - 0} < \epsilon$
This is the definition of the limits at infinity of the real exponential.
The result follows from Exponential Tends to Zero and Infinity.
$\blacksquare$