Complex Numbers - Basic Definitions
by M. Bourne
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Examples of quadratic equations:
- `2x^2 + 3x − 5 = 0`
- `x^2 − x − 6 = 0`
- `x^2 = 4`
The roots of an equation are the x-values that make it "work" We can find the roots of a quadratic equation either by using the quadratic formula or by factoring.
We can have 3 situations when solving quadratic equations.
Case 1: Two roots
Example: `2x^2 + 3x − 5 = 0`
We proceed to solve this equation using the quadratic formula as we did earlier:
`=(-3-sqrt(49))/4 or (-3+sqrt49)/4`
`=-2.5 or 1`
We have found 2 roots.
The graph of the quadratic equation ` y = 2x^2 + 3x − 5` cuts the `x`-axis at `x = -2.5` and `x = 1`, as expected, showing our 2 roots:
More examples of quadratic equations with 2 roots:
`x^2 = 4` has 2 solutions, `x = -2` and `x = 2`.
`x^2 − x − 6 = 0` has 2 solutions, `x = -2` and `x = 3`.
`2x^2 + 13x − 7 = 0` has 2 solutions, `x = -7` and `x = 1/2`.
Case 2: One Root
Example: `4x^2 − 12x + 9 = 0`
Notice what happens when we use the quadratic formula this time. Under the square root we get `144 − 144 = 0`.
So it means we only have one root. We can also say that this is a repeated root, since we are expecting 2 roots.
On the graph of `y = 4x^2 −12x + 9`, we can see that the graph cuts the x-axis in one place only, at `x = 1.5`.
Case 3: No Real Roots
Example: `x^2 −4x + 20 = 0`
This example gives us a problem. Under the square root, we get sqrt(-64), and we have been told repeatedly by our teachers that we cannot have the square root of a negative number. Can we find such a root?
A quadratic equation has degree 2 (the highest power of x is 2) and we can have either 2 real roots, one real repeated root or something that involves the square root of a negative number.
Cubic equations are polynomials which have degree 3 (this highest power of x is 3).
In the case of a cubic equation, we expect (up to) 3 real solutions:
Example 1: `x^3 − 2x^2 − 5x + 6 = 0` has solutions `x = -2, 1 and 3`.
Example 2: If `x^3 = 8`, we know the solution `x = 2`, but we expect 2 other solutions. What are they?
To allow for these "hidden roots", around the year 1800, the concept of
was proposed and is now accepted as an extension of the real number system. The symbol used is
`j = sqrt(-1)`
and `j` is called an imaginary number.
Why Not i for Imaginary Numbers?
Many textbooks use `i` as the symbol for imaginary numbers. We use `j`, because the main application of imaginary numbers is in electricity and electronics, so there is less confusion with `i` (which is used for current).
Your calculator or computer algebra system will probably use `i`.
Powers of j
You may need to look at this reminder example about multiplying square roots before you go any further.
`(sqrta)^2 = a`, for any value of `a`.
`j = sqrt(-1)`
Using these, we can derive the following:
`j^2 = (sqrt-1)^2 = -1`
Multiplying by `j` again gives us:
`j^3 = j^2(j) = -j`
Continuing the process gives us:
`j^4 = j^3(j) = -j(j) = -(-1) = 1`
`j^5 = j^4(j) = 1 × j = j`
`j^6 = j^5(j) = j × j = -1` etc
Example 3: Using `j`
Express the following in terms of the imaginary number `j`:
e. `sqrt(-2 × -18)`
(NOT the same as Number 4! - Note the difference.)
Complex numbers have a real part and an imaginary part.
Example 4: Complex numbers
a. `5 + 6j`
Real part: `5`, Imaginary part: `6j`
b. `−3 + 7j`
Real part: ` −3`, Imaginary part: `7j`
We can write the complex number `2 + 5j` as `2 + j5`.
There is no difference in meaning.
Solving Equations with Complex Numbers
We now return to our problem from above. We didn't know then what to do with `sqrt(-64)`. Now we can write the solution using complex numbers, as follows:
`=2-4j or 2+4j`
Equivalent Complex Numbers
Two complex numbers `x + yj` and `a + bj` are equivalent if:
The real parts are equal (`x = a`), and
The imaginary parts are equal (`y = b`).
Example 5: Equivalent complex numbers
Given that `3 + 2j= a + bj`, then
`a = 3` and `b = 2`.
1. Express in terms of `j`:
2. Simplify each of the following:
c. `j^2 − j^6`
Forms of Complex Numbers
We can write complex numbers in 3 different ways:
|Rectangular form:||`x + yj`||`5 + 6j`|
|Polar form:||`r(cos\ θ + j\ sin\ θ)`||`8(cos\ 24^"o" + j\ sin\ 24^"o")`|
|Exponential form:||`r\ e^(\ j\ θ)`||`6\ e^(2.5j)`|