Complex Exponential Tends to Zero

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Theorem

Let $\exp z$ be the complex exponential function.

Then:

$\displaystyle \lim_{\map \Re z \mathop \to +\infty} e^{-z} = 0$

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:

\(\displaystyle \size {e^{-z} - 0}\) \(=\) \(\displaystyle \size {e^{-z} }\)
\(\displaystyle \) \(=\) \(\displaystyle \size {e^{-x} }\) Modulus of Exponential is Modulus of Real Part
\(\displaystyle \) \(=\) \(\displaystyle \size {e^{-x} - 0}\)

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.

$\blacksquare$