Complex Sequence is Null iff Modulus of Sequence is Null/Proof 1

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

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 $n \ge N$.

This article, or a section of it, needs explaining. In particular: What does $\cmod {\paren {\cmod {z_n} } }$ mean? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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

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

This article, or a section of it, needs explaining. In particular: How does the above equation follow from Complex Modulus of Real Number equals Absolute Value? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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:

$\cmod {z_n} \to 0$

$\blacksquare$