19 - Complex Sequences

Convergence of a Complex Sequence

Let $(z_n)$ be a complex sequence
Let $L\in \mathbb{C}$
Then $(z_n)$ converges to $L$ as $n\to \infty$ if

$$\forall ϵ>0:\exists N\in \mathbb{N}\text{ such that }\forall n\ge N:|z_n-L|<ϵ$$

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Bolzano-Weierstrass Theorem for Complex Sequences corollary

Let $(z_n)$ be a bounded complex sequence then

$$(z_n)\text{has a convergent subsequence}$$

Proof

Write $z_n=x_n+i{y}_n$ where $x_n,y_n\in \mathbb{R}$
Suppose $(z_n)$ is bounded by $M$ so tthat

$$|z_n|\le M\text{ for all }n$$

Then $(x_n)$ and $(y_n)$ are also bounded by $M$, and are real sequences
By Bolzano-Weierstrass Theorem then

$$(x_n)\text{ has a convergent subsequence }(x_{n_r}{)}_r$$

Now $(y_{n_r}{)}_r$ is a bounded real sequence so by Bolzano-Weierstrass Theorem

$$(y_{n_r}{)}_r\text{ has a convergent subsequence }(y_{n_{r_s}}{)}_s$$

As $(x_{n_{r_s}}{)}_s$ is a subsequence of $(x_{n_r}{)}_r$ then it converges
Hence by Convergence of Complex Sequences then $(z_{n_{r_s}}{)}_s$ converges