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

Proof 1
By the definition of convergent sequence, we have:


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

Similarly:


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


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


 * $z_n \to 0$




 * $\cmod {z_n} \to 0$

Proof 2
Observe:

Therefore: