Complex Sequences

The definition of convergence of complex-valued sequences is a straightforward generalizations of that of real sequences.

Definition. Let $(z_n)$ is a sequence of complex numbers. Then $(z_n)$ converges to $z$ or $z_n \to z$ if

\[(\forall \varepsilon > 0)(\exists N \in \mathbb{N})(\forall n \ge N)\;|z_n - z| < \varepsilon\]

Definition. $(z_n)$ is a Cauchy sequence if

\[(\forall \varepsilon > 0)(\exists N \in \mathbb{N})(\forall m, n \ge N)\; |z_m - z_n| < \varepsilon\]

The following elementary result associates complex sequences with real sequences.

Proposition. Suppose that $(z_n = x_n + i y_n)$ is a sequence in $\mathbb{C}$ and $z = x + iy \in \mathbb{C}$. Then $z_n \to z$ iff $x_n \to x$ and $y_n \to y$. $(z_n)$ is a Cauchy sequence iff $(x_n)$ and $(y_n)$ are Cauchy sequences.

Proof.

($\Rightarrow$) Since $\vert x_n - x \vert \le \vert z_n - z \vert$ and $\vert y_n - y \vert \le \vert z_n - z \vert$. If $z_n \to z$, then $x_n \to x$ and $y_n \to y$.

($\Leftarrow$) Since $\vert z_n - z \vert \le \vert x_n - x \vert + \vert y_n - y \vert$. If $x_n \to x$ and $y_n \to y$, then $z_n \to z$.

The proof for the result concerning Cauchy sequences is essentially the same.

Definition. A sequence $(z_n)$ is bounded if the set $\Set{\vert z_n \vert : n \in \mathbb{N}}$ is bounded in $\mathbb{R}$.

With the above results, all the properties such as addition and multiplication of real sequences is applicable to complex sequences.

Proposition. [Bolzano-Weierstrass Theorem] Suppose $(z_n)$ is a bounded sequence of complex numbers. Then there is a subsequence $(z_{n_k})$ which converges.

Proof.

By real Bolzano-Weierstrass Theorem, there exists a subsequence $(z_{m_k})$ such that $(x_{m_k})$ converges and a subsequence $(z_{n_k})$ such that $(y_{n_k})$ converges. Hence, $(z_{n_k})$ converges.

Proposition. [General Principle of Convergence] A sequence $(z_n)$ of complex numbers is convergent iff it is a Cauchy sequence.

Proposition. Let $z_n = z^n$. Then $z_n \to 0$ if $\vert z \vert < 1$, $z_n \to 1$ if $z = 1$, otherwise it diverges.

Proof.

We have $\vert z_n - 0 \vert = \vert z \vert^n$.

If $\vert z \vert < 1$, $\vert z \vert^n \to 0$, therefore $z_n \to 0$.

Obviously, if $z = 1$, $z_n = 1$ and therefore $z_n \to 1$.

If $\vert z \vert \ge 1$ and $z \not= 1$, then $\vert z_{n+1} - z_n \vert = \vert z^n \vert \vert z - 1 \vert \ge \vert z - 1 \vert$. So it is not a Cauchy sequence and diverges.

References