Sequence Converges to Within Half Limit/Complex Numbers

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Theorem

Let $\sequence {z_n}$ be a sequence in $\C$.

Let $\sequence {z_n}$ be convergent to the limit $l$.

That is, let $\displaystyle \lim_{n \mathop \to \infty} z_n = l$ where $l \ne 0$.


Then:

$\exists N: \forall n > N: \cmod {z_n} > \dfrac {\cmod l} 2$


Proof

Suppose $l > 0$.

Let us choose $N$ such that:

$\forall n > N: \cmod {z_n - l} < \dfrac {\cmod l} 2$


Then:

\(\displaystyle \cmod {z_n - l}\) \(<\) \(\displaystyle \frac {\cmod l} 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \cmod l - \cmod {z_n}\) \(\le\) \(\displaystyle \cmod {z_n - l}\) Reverse Triangle Inequality
\(\displaystyle \) \(<\) \(\displaystyle \frac {\cmod l} 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \cmod {z_n}\) \(>\) \(\displaystyle \cmod l - \frac {\cmod l} 2\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\cmod l} 2\)

$\blacksquare$