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# Performing Arithmetic with Complex Numbers

Concept

## The Imaginary Unit

Is there any number that, when multiplied by itself, equals $\text{-} 1?$ Among the real numbers, the answer is no. However, the system of numbers can be expanded to include the imaginary unit $i,$ defined as $i^2 = \text{-} 1.$ From this definition, it follows that $\sqrt{\text{-} 1}=i,$ which allows the square root of any negative number to be found. What results is called an imaginary number. Once $i$ replaces the negative sign, the square root of the remaining positive number can be evaluated as usual.

$\sqrt{\text{-} a} = \sqrt{a} \cdot \sqrt{\text{-} 1} = \sqrt{a} \cdot i$
$\ \quad$ Condition: $a > 0$
Concept

## Complex Numbers

The set of complex numbers, represented by the symbol $\mathbb{C},$ is formed by all numbers that can be written in the form $z=a+bi,$ where $a$ and $b$ are real numbers, and $i$ is the imaginary unit. Here, $a$ is called the real part and $b$ is called the imaginary part of the complex number.

If $a=0,$ the number is an imaginary number. Conversely, if $b=0,$ the number is real. Both real numbers and imaginary numbers are subsets of the complex number set.
Method

## Performing Operations with Complex Numbers

Complex numbers can be added, subtracted and multiplied just like real numbers.

Method

Adding and subtracting complex numbers is done by combining like terms. Imaginary numbers and real numbers are not like terms.

Method

### Multiplication

When multiplying two complex numbers, the Distributive Property can be used. Meaning, each term of the first number is multiplied by each term of the second number.
Exercise

Simplify the following expressions. $\sqrt{\text{-} 36} \qquad (8i)^2 \qquad (2+4i)+(1-i) \qquad (1-i)(2+4i)$

Solution

Let's simplify the expressions, one at a time.

Example

### $\sqrt{\text{-} 36}$

The square root of a negative number can be found by first writing the number as a product, then using $\sqrt{\text{-} 1}=i.$
$\sqrt{\text{-} 36}$
$\sqrt{36} \cdot i$
$6i$
Example

### $(8i)^2$

Here, we can use the Power of a Product Property and that $i^2=\text{-} 1$, to simplify the expression.
$(8i)^2$
$64i^2$
$64(\text{-} 1)$
$\text{-}64$
Example

### $(2+4i)+(1-i)$

For this expression, we'll combine the real and the imaginary parts.
$(2+4i)+(1-i)$
$2+4i+1-i$
$2+1+4i-i$
$3+3i$
Example

### $(1-i)(2+4i)$

Here, we can use the Distributive Property to multiply two complex numbers. We'll also use the fact that $i^2=\text{-} 1.$
$(1-i)(2+4i)$
$1\cdot2+1\cdot4i-i\cdot2-i\cdot4i$
$2+4i-2i-4i^2$
$2+4i-2i-4(\text{-}1)$
$2+4i-2i+4$
$2+4+4i-2i$
$6+2i$
The answers are, in order: $6i \qquad \text{-}64 \qquad 3+3i\qquad 6+2i.$
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