Sequences of Complex Numbers

Sequences of Complex Numbers

A sequence of real numbers is an infinite ordered list $(a_n)_{n=1}^{\infty}$ of real numbers. We can similarly define a sequence of complex numbers.

Definition: A Sequence of Complex Numbers is an infinite ordered list of complex numbers $(z_n)_{n=1}^{\infty} = (z_1, z_2, ..., z_n, ...)$ where $z_n \in \mathbb{C}$ for each $n \in \mathbb{N}$. For each $n \in \mathbb{N}$, $z_n$ is called a $n^{\mathrm{th}}$ Term of the sequence.

It is convenient to denote the first term of a sequence by $z_1$. We can start a sequence at any index though. This will not cause any problems.

For example, the first few terms of the complex sequence $(n + i^n)_{n=1}^{\infty}$ are:

\begin{align} \quad (n + i^n)_{n=1}^{\infty} = (1 + i, 1, 3 - i, 5, 5 + i, ...) \end{align}

Many of the definitions familiar with sequences of real numbers are readily extended to sequences of complex numbers.

Definition: A sequence of complex numbers $(z_n)_{n=1}^{\infty}$ is said to Converge to $Z \in \mathbb{C}$ if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\mid z_n - Z \mid < \epsilon$. If the sequence $(z_n)_{n=1}^{\infty}$ does not converge to any $Z \in \mathbb{C}$ then $(z_n)_{n=1}^{\infty}$ is said to Diverge.

For example, consider the sequence $\left ( \frac{i^n}{n} \right)_{n=1}^{\infty}$. The first few terms of this sequence are:

\begin{align} \quad \left ( \frac{i^n}{n} \right)_{n=1}^{\infty} = \left ( i, \frac{-1}{2}, \frac{-i}{3}, \frac{1}{4}, ... \right ) \end{align}

We claim that this sequence converges to $Z = 0$. To prove this, let $\epsilon > 0$ be given. Then we want $\biggr \lvert \frac{i^n}{n} - 0 \biggr \rvert < \epsilon$. Now:

\begin{align} \quad \biggr \lvert \frac{i^n}{n} \biggr \rvert = \frac{\mid i^n \mid}{\mid n \mid} = \frac{1}{n} \end{align}

For any given $\epsilon > 0$, choose $N \in \mathbb{N}$ such that $\displaystyle{N > \frac{1}{\epsilon}}$. Then if $n \geq N$ we have that:

\begin{align} \quad \frac{1}{n} \leq \frac{1}{N} < \epsilon \end{align}

In other words, for $n \geq N$:

\begin{align} \quad \biggr \lvert \frac{i^n}{n} \biggr \rvert = \frac{1}{n} \leq \frac{1}{N} < \epsilon \end{align}

Therefore the sequence $\left ( \frac{i^n}{n} \right )_{n=1}^{\infty}$ converges to $Z = 0$.

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