Complex Exponential Tends to Zero

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Theorem

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

Then:

$\displaystyle \lim_{\operatorname{Re} \left({z}\right) \mathop \to +\infty} e^{-z} = 0$

where $\operatorname{Re} \left({z}\right)$ 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:

$\displaystyle x > M \implies \left \vert {e^{-z} - 0} \right\vert < \epsilon$

But:

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

so we need an $M$ such that:

$\displaystyle x > M \implies \left \vert {e^{-x} - 0} \right\vert < \epsilon$

This is the definition of the limit at infinity of the real exponential.

The result follows from Exponential Tends to Zero and Infinity.

$\blacksquare$