Complex Sequence is Null iff Modulus of Sequence is Null

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Theorem

Let $\sequence {z_n}_{n \mathop \in \N}$ be a complex sequence.


Then:

$z_n \to 0$

if and only if:

$\cmod {z_n} \to 0$

That is:

$\sequence {z_n}_{n \mathop \in \N}$ is a null sequence if and only if $\sequence {\cmod {z_n} }_{n \mathop \in \N}$ is a null sequence.


Proof 1

By the definition of convergent sequence, we have:

$z_n \to 0$ if and only if for each $\epsilon > 0$ there exists $N \in \N$ such that $\cmod {z_n} < \epsilon$ for each $n \ge N$.

Similarly:

$\cmod {z_n} \to 0$ if and only if for each $\epsilon > 0$ there exists $N \in \N$ such that $\cmod {\paren {\cmod {z_n} } } < \epsilon$ for each $n \ge N$.



From Complex Modulus of Real Number equals Absolute Value, we have:

$\cmod {\paren {\cmod {z_n} } } = \cmod {z_n}$



So:

$\cmod {z_n} \to 0$ if and only if for each $\epsilon > 0$ there exists $N \in \N$ such that $\cmod {z_n} < \epsilon$ for each $n \ge N$.

We can therefore see that:

$z_n \to 0$

if and only if:

$\cmod {z_n} \to 0$

$\blacksquare$


Proof 2

Observe:

\(\text {(1)}: \quad\) \(\ds \cmod {z_n - 0}\) \(=\) \(\ds \cmod {z_n}\)
\(\ds \) \(=\) \(\ds \cmod {z_n} - 0\)
\(\ds \) \(=\) \(\ds \cmod {\cmod {z_n} - 0}\)

Therefore:

\(\ds \lim_{n \to \infty} z_n\) \(=\) \(\ds 0\)
\(\ds \leadstoandfrom \ \ \) \(\ds \forall \epsilon \in \R_{>0}: \exists N \in \R: \, \) \(\ds n > N\) \(\implies\) \(\ds \cmod {z_n - 0} < \epsilon\) Definition of Convergent Complex Sequence
\(\ds \leadstoandfrom \ \ \) \(\ds \forall \epsilon \in \R_{>0}: \exists N \in \R: \, \) \(\ds n > N\) \(\implies\) \(\ds \cmod {\cmod {z_n} - 0} < \epsilon\) from $\paren 1$
\(\ds \leadstoandfrom \ \ \) \(\ds \lim _{n \to \infty} \cmod {z_n}\) \(=\) \(\ds 0\) Definition of Convergent Complex Sequence

$\blacksquare$