Complex Numbers

Because the square of any real number is non-negative, a simple equation such as x² = -4 has no real solutions.

To handle this situation, we look at a larger set of numbers called complex numbers.

We write complex numbers in terms of the imaginary number i. This is defined as the square root of -1:

i = √-1       or             i ² = -1

Using this, we define a complex number as:

a + bi

where a and b are real numbers.

The form a + bi is called standard form of a complex number.

The number a is called the real part of the complex number.

The number b is called the imaginary part of the complex number.

For example,

The number 7 + 5i has a real part of 7 and an imaginary part of 5.

The number -4 – 3i has a real part of -4 and an imaginary part of -3.

Adding and Subtracting Complex Numbers

To add or subtract imaginary numbers we deal with the real parts and the imaginary parts separately. We basically add or subtract like terms.

Example 1: Add

a)      (4 +3i) + (5 + 9i)                     b) (-6 + 4i) + (8 – 7i)

c)  (1/2 + 3/4 i) + (2/3 + 1/5 i)

Example 2:  Subtract

a)      (9 + 8i) – (5 + 3i)

b)      (3 – 2i) – (4 – 10i)

Simplifying Square Roots of Negative Numbers:

Using the definition i = √-1, we can simplify square roots of negative numbers.

Forexample,

√-4 = √-1√4=i√4 = 2i

√-17 = √-1√17 = i√17

√-24 = √-1√24 = i√24 = i√4√6 = 2i√6

Multiplying and Dividing Complex Numbers

To multiply radicals of radical numbers we must use the following order of operations:

1.      First, change each radical to a complex number, so write all expressions of the form √-b as i√b.

2.      Then multiply the i factors together and the numbers inside the radicals together.

3.      Simplify using i² = -1 and reduce radicals to lowest form

Example: Simplify

a)      √-5√-7

b)      √-2√-8

c)      √-6√-8

d)     √-75/√-3

e)      √-48/√12

Multiplying complex numbers

Complex numbers are usual in binomial form a + ib, so we can find the product of two complex numbers the same way as we find the product of two binomial, using distribution or FOIL.

Example: Multiply and simplify

a)      (2 + 3i)(4 + 5i)

b)      (-3 + 6i)(2 – 4i)

c)      (1 – 7i)²

d)     (2 + 3i)(2 – 3i)

Dividing complex numbers:

Notice that when we multiply the sum and difference of numbers the middle terms cancel out. For example:

(2 + 3i)(2 – 3i) = 4 – 6i + 6i - 3i²

 The middle terms cancel out and we get

= 4 + 3i²

= 4 – 3 = 1

Two complex numbers a + bi and a – bi are called complex conjugates of each other.

We multiply a number by its conjugate if we want to eliminate the imaginary parts as we saw in the example above.

When we divide complex numbers, we need to get rid of the i in the denominator in order to write it in standard form, so we need to multiply the top and the bottom of the fraction by the complex conjugate of the denominator.

Example: Express the following numbers in standard form.

a)      3i/(5 + 2i)

b)      (2 – 3i)/(4 – 7i)

c)      (4 – 5i)/2i

Now try this Practice Quiz