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:
The curve y = 2x2 + 3x − 5, showing x-intercepts at (−2.5, 0) and (1, 0).
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`.
The curve y = 4x2 − 12x + 9, showing x-intercept at (1.5, 0).
Case 3: No Real Roots
Example: `x^2 −4x + 20 = 0`
This example gives us a problem. Under the square root, we get `(-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?
The curve y = x2 − 4x + 20, which has no x-intercepts.
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`.
The curve y = x3 − 2x2 − 5x + 6, which has 3 x-intercepts.
Example 2: If `x^3 = 8`, we know the solution `x = 2`, but we expect 2 other solutions. What are they?
The curve y = x3 − 8, which only has one x-intercept.
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)`|