by M. Bourne
On this page:
- Roots of cubic equations
- Imaginary Numbers
- Powers of j
- Complex Numbers
- Equivalent complex numbers
- Forms of a complex number
Let's first consider what we learned before in Quadratic Equations and Equations of Higher Degree, so we can better understand where complex numbers are coming from.
Quadratic Equations
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:
`x=(-b+-sqrt(b^2-4ac))/(2a)`
`=(-3+-sqrt(9+40))/4`
`=(-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 2solutions, `x = -2` and `x = 2`.
`x^2 − x − 6 = 0` has2 solutions, `x = -2` and `x = 3`.
`2x^2 + 13x − 7 = 0` has2 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`.
`x=(-b+-sqrt(b^2-4ac))/(2a)`
` =(12+-sqrt(144-144))/8`
`=12/8`
`=1.5`
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`
`x =(-b+-sqrt(b^2-4ac))/(2a)`
`=(4+-sqrt(16-80))/2`
`=(4+-sqrt(-64))/2`
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.
Summary
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
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) 3real 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.
Imaginary Numbers
To allow for these "hidden roots", around the year 1800, the concept of
`sqrt(-1)`
was proposed and is now accepted as an extension of the realnumber 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 andelectronics, 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 reminderexample about multiplying square roots before you go anyfurther.
Reminder example
Reminder Example about Repeated Multiplying of Square Roots
If `x = sqrt10`, then
`x^2 = (sqrt10)^2 = 10`
[This because square and square root are inverse processes.]
Let's multiply our previous answer by `sqrt10`.
`x^3 = (sqrt10)^3 = (sqrt10)^2sqrt10` ` = 10sqrt10`
The next step in the pattern is:
`x^4 = (sqrt10)^4 = 10 *(sqrt10)^2` ` = 10 × 10 = 100`
[Once again, we are multiplying our previous answer by `sqrt10`.]
The next step in the pattern is:
`x^5 = (sqrt10)^5 = 100sqrt10`
The next step in the pattern is:
`x^6 = (sqrt10)^6 = 1000`
What we are doing next in Powers of j works in the same way.
Recall:
`(sqrta)^2 = a`, for any value of `a`.
and
`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`:
a. `sqrt(-16)`
Answer
`sqrt(-16)`
`sqrt(-1)sqrt(16)=jxx4=4j`
b. `sqrt(-100)`
Answer
`sqrt(-100)`
`sqrt(-1)sqrt(100)=jxx10=10j`
c. `sqrt(-7)`
Answer
`sqrt(-7)`
`sqrt(-1)sqrt(7)=jxxsqrt(7)=jsqrt(7)`
Care: We don't write this as `sqrt(7)j`, because it could be confused with `sqrt((7j)`
d. `sqrt(-2)sqrt(-18)`
Answer
`sqrt(-2)sqrt(-18)`
`=jsqrt(2)xxjsqrt(18)`
`=j^2xxsqrt(2xx18)`
`=(-1)sqrt(36)`
`=-6`
e. `sqrt(-2 × -18)`
(NOT the same as Number 4! - Notethe difference.)
Answer
`sqrt((-2)(-18))`
`sqrt(36)=6`
Complex Numbers
Complex numbers have a real part and animaginary part.
Example 4: Complex numbers
a. `5 + 6j`
Real part: `5`, Imaginary part: `6j`
b. `−3 + 7j`
Real part: ` −3`, Imaginary part: `7j`
Notation
We can write the complex number `2 +5j` as `2 + j5`.
There is no difference inmeaning.
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:
`x=(4+-sqrt(-64))/2`
`=(4+-jsqrt(64))/2`
`=(4+-8j)/2`
`=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`.
Exercises
1. Express in terms of `j`:
`-sqrt(-2/5)`
Answer
`-sqrt(-2/5) `
`= -sqrt((-1)(2/5))`
` = -sqrt(-1)xxsqrt(2/5)`
`= -jsqrt(2/5)`
2. Simplify each of the following:
a. `sqrt(-2)sqrt(-8)`
Answer
`sqrt(-2)sqrt(-8)`
`=(sqrt(-1)sqrt2)(sqrt(-1)sqrt8)`
`=(jsqrt2)(jsqrt8)`
`=j^2sqrt16`
`=(-1)(4)`
`=-4`
b. `sqrt((-2)(-8))`
Answer
`sqrt((-2)(-8))=sqrt16=4`
c. `j^2 − j^6`
Answer
`j^2 = -1`, and
`j^6 = (j^2)^3 = (-1)^3 = -1`
So
`j^2 − j^6` `=(-1) − (-1) = 0`
d. `(sqrt(-2))^2+j^4`
Answer
`(sqrt(-2))^2 = -2`, and
`j^4 = (j^2)^2 = (-1)^2 = 1`, so
`(sqrt(-2))^2+j^4` `=-2 + 1 = -1`
Forms of Complex Numbers
We can write complex numbers in 3 different ways:
We can write complex numbers in 3 different ways:
Rectangular form: x + yj
Example: 5 + 6j
Polar form: r(cos θ + j sin θ)
Example: 8(cos24° + j sin 24°)
Exponential form: rejθ
Example: 6e2.5j
We will meet polar form and exponential form later in thischapter, but first, let's see how to perform basic operations with complex numbers.
