30
COMPLEX NUMBERS
The real and imaginary components
IN THE LESSON ON RADICALS, we saw how to solve any equation in this form:
| x² | = | a. | |
| The solution is | |||
| x | = | ± . |
|
| If we apply that rule to this equation -- | |||
| x² | = | −1 | |
| -- then | |||
| x | = | ± . |
|
But is not a real number. There is no positive or negative number whose square will be negative. Nevertheless, it turns out to be extremely useful in mathematics and science to say that the equation
x² + 1 = 0
has a solution.
is called a complex number or an imaginary number. It is the imaginary unit. Its symbol is i.
i = .
The complex number i is purely algebraic. That is, we call it a "number" because it will obey all the rules we normally associate with a number. We may add it, subtract it, multiply it, and so on.
The basic algebraic property of i is the following:
i² = −1
Example 1. 3i· 4i = 12i² = 12(−1) = −12.
Example 2. −5i· 6i = −30i² = 30.
We can see, then, that the factor i² changes the sign of a product.
Problem 1. Evaluate the following.
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| a) | i² = −1 | b) | i· 2i = 2i² = 2(−1) = −2 | |
| c) | (3i)² = 3²i² = −9 | d) | −5i· 4i = −20i² = 20 | |
Negative radicand
If a radicand is negative --
, where a > 0,
-- then we can simplify it as follows:
= 
= = i.
| Examples 3. | 
| = | i |
| = | i = 2i |
|
| = | i = 2i |
|
Problem 2. Express each of the following in terms of i.
| a) | = i |
b) |  = 3i |
c) |  = 7i |
|||
| d) |  = i |
e) |  = i |
f) |  = 2i |
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| g) |  = 3i |
h) |  = 5i |
i) |  = 7i |
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Powers of i
Let us begin with i0, which is 1. (Any number with exponent 0 is 1.) Each power of i can be obtained from the previous power by multiplying it by i. We have:
| i0 | = | 1 |
| i1 | = | i |
| i2 | = | −1 |
| i3 | = | −1· i = −i |
| i4 | = | −i· i = −i² = −(−1) = 1 |
And we are back at 1 -- the cycle of powers will repeat! Any power of i will be either
1, i, −1, or −i
-- according to the remainder upon dividing the exponent n by 4.
Examples 4 .
| i9 | = | i, because on dividing 9 by 4, the remainder is 1. i9 = i1. | |
| i18 | = | −1, because on dividing 18 by 4, the remainder is 2. i18 = i2. | |
| i35 | = | −i, because on dividing 35 by 4, the remainder is 3. i35 = i3. | |
| i40 | = | 1, because on dividing 40 by 4, the remainder is 0. i40 = i0. | |
Note: Even powers of i will be either 1 or −1, according as the exponent is a multiple of 4 or 2 more than a multiple of 4. While odd powers will be either i or −i.
Problem 3. Evaluate each power of i.
| a) i3 = −i | b) i4 = 1 | c) i6 = i2 = −1 | ||
| d) i9 = i1 = i | e) i12 = i0 = 1 | f) i17 = i1 = i | ||
| g) i27 = i3 = −i | h) i30 = i2 = −1 | i) i100 = i0 = 1 | ||
Algebra with complex numbers
Complex numbers follow the same rules as real numbers. For example, to multiply
(2 + 3i)(2 − 3i)
the student should recognize the form (a + b)(a − b) -- which will produce the difference of two squares. Therefore,
| (2 + 3i)(2 − 3i) | = | 4 − 9i² |
| = | 4 − 9(−1) | |
| = | 4 + 9 | |
| = | 13. | |
Again, the factor i² changes the sign of the term.
Problem 4. Multiply.
a) (1 + i
)(1 − i) =
1 − 2i²
=
1 + 2 = 3
| b) (3 − i)² |
= | 9 − 6i + 2i², |
upon squaring the binomial, |
| = | 9 − 6i − 2 |
||
| = | 7 − 6i |
||
| c) (2 + 3i)(4 − 5i) | = | 8 − 10i + 12i − 15i² |
| = | 8 + 2i + 15 | |
| = | 23 + 2i | |
Problem 5. (x + 1 + 3i)(x + 1 − 3i)
a) What form will that produce? The difference of two squares.
b) Multiply out.
| (x + 1 + 3i)(x + 1 − 3i) | = | (x + 1)² − 9i² |
| = | x² + 2x + 1 + 9 | |
| = | x² + 2x + 10 | |
| c) | (x − 2 − i)(x − 2 + i) |
= | (x − 2)² − 2i² |
| = | x² − 4x + 4 + 2 | ||
| = | x² − 4x + 6 | ||
The real and imaginary components
Here is the standard form of a complex number:
a + bi,
where both a and b are real. For example,
3 + 2i.
a -- that is, 3 in the example -- is called the real component (or the real part). b (2 in the example) is called the imaginary component (or the imaginary part). Again, the components are real.
Problem 6. Name the real component a and the imaginary component b.
| a) | 3 − 5i a = 3, b = −5. | b) | 1 + i a = 1, b = . |
|
| c) | i a = 0, b = 1. | d) | −6 a = −6, b = 0. | |
Complex conjugates
The complex conjugate of a + bi is a − bi. The main point about a conjugate pair is that when they are multiplied --
(a + bi)(a − bi)
-- a positive real number is produced. For, that form is the difference of two squares:
(a + bi)(a − bi) = a² − b²i² = a² + b²
The product of a conjugate pair
is equal to the sum of the squares of the components.
Problem 7. Calculate the positive real number that results from multiplying each number with its complex conjugate.
a) 2 + 3i. (2 + 3i)(2 − 3i) = 2² + 3² = 4 + 9 = 13
b) 3 − i.
(3 − i)(3 + i) =
= 3² + ()²= 9 + 2 = 11
c) u + iv. (u + iv)(u − iv) = u² + v²
d) 1 + i. (1 + i)(1 − i) = 1² + 1² = 2
e) −i. (−i)(i) = −i² = 1
Next Lesson: Rectangular coordinates
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