COMPLEX NUMBERS AND COMPLEX SOLUTIONS

Objective:

• Write the square of a negative number in i-form and perform operations on numbers in i -form.
• Determine the equality of two complex numbers.
• Add, subtract and multiply complex numbers.
• Use complex conjugates to divide complex numbers.
Many quadratic equations do not have real number solutions.    In order to find their solutions, we must extend our usual   understanding of number to include a more general notion of   number.

Recall 1: The equation

x^2 - x + 1 = 0

has no real solutions.  If we attempt to use the quadratic   formula to find the solutions we get

Simplifying, we get x= [1+ sqrt (-3)]/2.  But what does  that mean?
There is no real number whose square is negative!  So, what do we do?  Extend the real numbers to a new category of   numbers called the complex numbers; these are numbers   of the form a + ib, where a and b are real numbers and i symbolically represents sqrt(-1).  In this way, our solution is  two complex numbers

1/2 + i* sqrt(3)/2    and  1/2 - i* sqrt(3)/2 .

Note 1
The thing to notice in Recall 1 is that 1/2 + i* sqrt(3)/2 is a complex number because it is of the form

(real number) + (another real number)i.

Moreover, you should notice that real numbers are a special case of complex numbers, since any real number may be written in the form

(real number) + (0)i.

Note 2
By extending the notion of number to include complex number, we guarantee that any quadratic equation will have a solution, although it may not be a real one.

Recall 2: Complex numbers may be added together or subtracted from one another, the result being a new complex number defined as

(a + bi) ± (c + di) = (a ± c) + (b ± d)i.

For example,

(3 + 2i) + (5 - 3i) = 8 - i.

Recall 3: Complex numbers may be multiplied together, the result being a new complex number defined as

(a + bi)(c + di) = (ac - bd) + (ad + bc)i.

For example,

(3 + 2i)(5 - 3i) = 21 + i.

Note 3
The usual laws of arithmetic--such as, the commutative law, associative law and distributive law--hold for complex numbers, just as they do for real numbers.

Recall 4: (Terminology)

When a complex number is written in the form a + bi, then  a is called the real part and b is called the imaginary part.

The conjugate of the complex number a + bi is the  complex number a - ib.  In other words, to form the   conjugate of a complex number, just change the sign of the  imaginary part.

The product of any complex number with its conjugate is a  real number.  To see this, take the product

(a + bi)(a - bi)

and follow the rule (Recall 2) for taking products of   complex numbers.

(a + bi)(a - bi) = a^2 + b^2.

For example,  (3 + 2i)(3 - 2i) = 3^2 + 2^2 = 13.

Notation
The conjugate of a complex number a + bi is denoted as   In other words,
= a - bi.

Note 4
Solutions to quadratic equations always come in conjugate pairs.  This means that if a + bi is a solution, then so is a - bi.  In the particular cases where the solutions are real numbers, the conjugate is equal to itself.  For example, if we interpret a real number a to be a special case of the complex number a + 0i, then its conjugate is a - 0i.  But this is just a again.

Examples

Example 1. Represent the complex number

(4 - 6i)(3 + 2i).

in the form a + bi

Solution: We may use the rule for products of complex numbers given in Recall 3
(a + bi)(c + di) = (ac - bd) + (ad + bc)i.

Or, we could simply treat i as   sqrrt(-1) and multiply out these numbers.  The numbers simplify if we use the fact that i^2 = -1.  We have

(4 - 6i)(3 + 2i) = (4*3 - (-6)2) + (4*2 + (-6)*3)i= 24 - 10i.

Example 2. Find the complex number that is equivalent to

(3 - 5i)().

and represent it in the form a + bi.

Solution: When you multiply a complex number (a + bi) by its conjugate the result is the real number that is the sum of the squares of a and b.

(3 - 5i) = 9 + 25 = 34.

Example 3  Find the complex number that is equivalent to

(2-3i)/(3+2i)

and represent it as a + bi.

Solution: Multiply the fraction by (3-2i)/(3-2i).  This does not change the fraction in any way.  But we now see that the original fraction is the same as

[(2-3i)/(3+2i)][(3-2i)/(3-2i)]

After multiplying the numerators and denominators, this fraction simplifies to 0 - i, or just plain -i.

Example 4. Solve for x in the equation

x^2 - 2x + 10 = 0.

Solution: Using the quadratic formula, we find that

After evaluating the expression, we find that

x = 1 + 3i and x = 1 - 3i

are the two complex conjugate solutions.

EXERCISES

Write each of the complex numbers in the standard form a + bi:

1. (3 - 5i) + (3 - 5i)

2. (2 - 3i)(3 - 2i)

3. (2 - 3i)(3 + 2i)

4. (1 + 3i) + (1 - 3i)

5.

6  (3+3i) / (3-3i)

7  (4+5i)/(3-3i)

8 Find the solutions to the quadratic equation  x^2 + 9 = 0.

9 Find the solutions to the equation x^2 + x + 1 = 0.

10 Solve the quadratic equation  5x^2 - 2x + 3 = 0.

11 Solve the equation 6x^2 - 2x + 2 = 2x^2 - x - 3

12 Solve the equation 2x^2 + 3x + 2 = x - 3