# Complex Numbers Examples 1

Recall from the Complex Numbers page that numbers in the form $z = a + bi$ where $a, b \in \mathbb{R}$ and $i = \sqrt{-1}$ are called complex numbers and the set of all complex numbers is denoted by $\mathbb{C}$. We will now look at some examples regarding complex numbers.

## Example 1

**Graph the numbers $-2 - i$ and $2 + i$ in the complex plane.**

The graph of these two complex numbers is given below:

## Example 2

**Determine the values of $i^1$, $i^2$, … $i^n$ for $n \in \mathbb{N}$.**

We first have that $i^1 = i$. So $i^2 = \sqrt{(-1)}^2 = -1$, and multiplying by $i$ again gives us that $i^3 = -i$. Multiplying by $i$ once more gives us $-i^2 = -(-1) = 1$.

Its not hard to see that:

(1)## Example 3

**Verify that the sequence $\{ \frac{i^{2n}}{n} \}_{n=1}^{\infty}$ converges to zero.**

Note that $i^{2n} = -1$ if $n$ is odd and $i^{2n} = 1$ if $n$ is even. The first few terms of this sequence is $\left \{ -1, \frac{1}{2}, -\frac{1}{3}, \frac{1}{4}, ... \right \}$.

We note that the numerator of this sequence is bounded, that is $-1 ≤ i^{2n} ≤ 1$ for all $n \in \mathbb{N}$ and the denominator is unbounded as $\lim_{n \to \infty} n = \infty$ and so $\lim_{n \to \infty} \frac{i^{2n}}{n} = 0$.

## Example 4

**Simplify the product $(2i - 3i^2 + 4)(-i + 2)$. Is this a real number?**

When we expand this product we get that:

(2)So this number is not a real number. Alterantively, we could have simplified the product from that start by noting that $-3i^2 = 3$, and so $(2i - 3i^2 + 4)(-i + 2) = (2i + 7)(-i + 2)$ and thus:

(3)