COMPLEX NUMBERS AND COMPLEX SOLUTIONS
Objective:
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
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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