Complex Sequence is Null iff Modulus of Sequence is Null/Proof 2

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

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

$\blacksquare$