Imaginary and complex numbers

Complex numbers are ordinary numbers expanded to include imaginary numbers. A complex number is written as $a+bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit. Every real number is also a complex number - it just has an imaginary part of $0$.

For example:

• A complex number might be $2+3i$.
• A “typical” real number can be written as $2+0i$, which simplifies to $2$.

So the number $2$ is a complex number whose imaginary part is $0$.

In this chapter, you’ll explore how imaginary numbers work in algebra.

Imaginary numbers

Imaginary unit $i$

In algebra, the letter $i$ refers to the imaginary unit. It marks a term as imaginary and lets you keep track of the imaginary part of a complex number. The value of $i$ is defined as:

$$i=\sqrt{(-1)}$$

From this definition, you can find the square of $i$:

$$i^2=-1$$

In many problems, these facts will be given. Still, it’s important to be comfortable with the repeating pattern of powers of $i$.

Imaginary unit exponent cycle

Above, you saw the values of $i$ and $i^2$. This cycle shows how to find $i^n$ for any integer exponent $n$. The idea is simple: each time the exponent increases by 1, you multiply the previous value by $i$.

Starting from $i^2=-1$, the values repeat in a pattern:

If this feels unfamiliar, connect it back to what an exponent means: repeated multiplication.

$$i=\sqrt{(-1)}$$

so

$$i^2=i\ast i=\sqrt{(-1)}\ast \sqrt{(-1)}$$

When you multiply a square root by itself, the square roots cancel. Here, that leaves $i^2=-1$.

Next, use exponent rules: $i^3=i^2\ast i$. Since $i^2=-1$, multiply:

$$i^2\ast i=-1\ast i=-i$$

Now rewrite $i^4$ as $i^3\ast i$. Since $i^3=-i$:

$$i^4=i^3\ast i=-i\ast i=-i^2$$

That expression includes $i^2$, which you already know equals $-1$. Substitute:

$$i^4=-(i^2)=-(-1)=1$$

One more step shows the cycle restarting:

$i^5=i^4\ast i=1\ast i=i$.

From here, the same pattern repeats forever. On tests, you’ll often use this cycle to simplify higher powers of $i$ quickly.

Below is a written form of the same pattern:

• $i^2=-1$
• $i^3=-i$
• $i^4=1$
• $i^5=i$
• $i^6=-1$
• $i^7=-i$

Imaginary numbers in algebra

When you add or subtract complex numbers, treat $i$ like a variable. Combine the real parts (terms without $i$), and combine the imaginary parts (terms with $i$).

Example:

$$(2+3i)-(4-i)=-2+4i$$

Multiplication works like ordinary algebra as well. If you’re multiplying two binomials, treat $i$ like a variable and FOIL.

$$\begin{gathered} (2+3i)\ast (4-i) \\ 8-2i+12i-3i^2 \\ -3i^2+10i+8 \end{gathered}$$

You’re not finished until you simplify powers of $i$. Since $i^2=-1$:

$$\begin{gathered} -3(-1)+10i+8 \\ 3+10i+8 \\ 11+10i \end{gathered}$$

Division is more complicated. To divide by a complex number with both a real and imaginary part, you use a complex conjugate.

A complex conjugate is the same complex number, but with the sign of the imaginary part flipped. For example:

$$\begin{gathered} 2+3i \\ 2-3i \end{gathered}$$

To divide, you multiply by a form of 1 that uses the conjugate of the denominator:

• Find the complex conjugate of the denominator.
• Multiply the fraction by (conjugate)/(conjugate).

This works because anything divided by itself equals $1$, and multiplying by $1$ doesn’t change the value.

Setup:

$$\frac{(2+3i)}{(1-i)}\ast \frac{(1+i)}{(1+i)}$$

Now multiply the numerators and denominators (FOIL both):

$$\begin{array}{r}\frac{[(2+3i)\ast (1+i)]}{[(1-i)\ast (1+i)]}\\ \\ \frac{(2+2i+3i+3i^2)}{(1+i-i-i^2)}\\ \\ \frac{(2+2i+3i+3i^2)}{(1+i-i-i^2)}\\ \\ \frac{(3i^2+5i+2)}{(-i^2+1)}\end{array}$$

Finally, simplify using $i^2=-1$:

$$\begin{array}{r}\frac{(3i^2+5i+2)}{(-i^2+1)}\\ \\ \frac{(3(-1)+5i+2)}{(-(-1)+1)}\\ \\ \frac{(3(-1)+5i+2)}{(-(-1)+1)}\\ \\ \frac{(-1+5i)}{2}\end{array}$$

This is the final simplified answer. Notice that the denominator no longer contains $i$ - that’s the goal when dividing complex numbers.

Key points

Complex numbers
All numbers are complex numbers, composed of both a real and imaginary part.

Imaginary numbers
The portion of complex numbers containing the imaginary unit $i$. Most normal numbers have an imaginary value equal to $0$.

Imaginary unit $i$
This unit is treated as a variable that represents an imaginary term.

Imaginary unit exponent cycle
$i^2=-1,i^3=-i,i^4=1,i^5=i$
The cycle continues in this pattern for increasing exponents of $i$.

Addition & subtraction
Add and subtract terms with and without $i$ as if it were a variable like $x$.

Multiplication
Multiply complex numbers similar to algebra, treating $i$ as a variable and FOILing when necessary.

Complex conjugate
A copy of the original complex number with the sign of the imaginary part opposite that of the original.

Division
Divide by multiplying the original fraction by a fraction composed of the complex conjugate as the numerator and the denominator. Simplify the product to result in the final answer.