# Complex Sequence is Null iff Modulus of Sequence is Null

## Theorem

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

Then:

$z_n \to 0$
$\cmod {z_n} \to 0$

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

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:

$z_n \to 0$
$\cmod {z_n} \to 0$

$\blacksquare$