College Algebra

0% Complete
0 out of 33 Lessons
0% Complete
0 out of 4 Exams
Previous Lessons
Open Chapter Ch. 1: Graphs, Functions, and Models
Lesson #1 Graphing
Incomplete Assignment Study Questions for Lesson 1
Lesson #2 Graphing Functions
Incomplete Assignment Study Questions for Lesson 2
Lesson #3 Slope of Linear Functions
Incomplete Assignment Study Questions for Lesson 3
Lesson #4 Linear Equations and Modeling
Incomplete Assignment Study Questions for Lesson 4
Lesson #5 Linear Equations, Functions, Zeros and Applications
Incomplete Assignment Study Questions for Lesson 5
Lesson #6 Linear Inequalities
Incomplete Assignment Study Questions for Lesson 6
Open Chapter Ch. 2: More on Functions
Lesson #7 Analyzing Functions
Incomplete Assignment Study Questions for Lesson 7
Exam Exam 1
Lesson #8 Algebra for Functions
Incomplete Assignment Study Questions for Lesson 8
Lesson #9 The Composition of Functions
Incomplete Assignment Study Questions for Lesson 9
Lesson #10 Symmetry
Incomplete Assignment Study Questions for Lesson 10
Lesson #11 Transformations
Incomplete Assignment Study Questions for Lesson 11
Lesson #12 Equations of Variation
Incomplete Assignment Study Questions for Lesson 12
Open Chapter Ch. 3: Quadratic Functions and Equations; Inequalities
Lesson #13 Complex Numbers
Incomplete Assignment Study Questions for Lesson 13
Lesson #14 Quadratic Equations, Functions, Zeros and Models
Incomplete Assignment Study Questions for Lesson 14
Lesson #15 Analyzing Graphs of Quadratic Functions
Incomplete Assignment Study Questions for Lesson 15
Lesson #16 Rational and Radical Equations
Incomplete Assignment Study Questions for Lesson 16
Exam Midterm Exam
Lesson #17 Absolute Value Equations and Inequalities
Incomplete Assignment Study Questions for Lesson 17
Open Chapter Ch. 4: Polynomial Functions and Rational Functions
Lesson #18 Polynomial Functions and Models
Incomplete Assignment Study Questions for Lesson 18
Lesson #19 Graphing Polynomial Functions
Incomplete Assignment Study Questions for Lesson 19
Lesson #20 Polynomial Division
Incomplete Assignment Study Questions for Lesson 20
Lesson #21 The Zeros of Polynomial Functions
Incomplete Assignment Study Questions for Lesson 21
Lesson #22 Rational Functions
Incomplete Assignment Study Questions for Lesson 22
Lesson #23 Polynomial and Rational Inequalities
Incomplete Assignment Study Questions for Lesson 23
Open Chapter Ch. 5: Exponential Functions and Logarithmic Functions
Lesson #24 Inverse Functions
Incomplete Assignment Study Questions for Lesson 24
Lesson #25 Exponential Functions and Graphs
Incomplete Assignment Study Questions for Lesson 25
Lesson #26 Logarithmic Functions and Graphs
Incomplete Assignment Study Questions for Lesson 26
Exam Exam 3
Lesson #27 Properties of Logarithmic Functions
Incomplete Assignment Study Questions for Lesson 27
Lesson #28 Solving Exponential Equations and Logarithmic Equations
Incomplete Assignment Study Questions for Lesson 28
Lesson #29 Applications and Models: Growth and Decay; Compound Interest
Incomplete Assignment Study Questions for Lesson 29
Open Chapter Ch. 6: Systems of Equations
Lesson #30 Systems of Equations in Two Variables
Incomplete Assignment Study Questions for Lesson 30
Lesson #31 Systems of Inequalities and Linear Programming
Incomplete Assignment Study Questions for Lesson 31
Lesson #32 Nonlinear Systems of Equations and Inequalities
Incomplete Assignment Study Questions for Lesson 32
Lesson #33 Sequences and Series
Incomplete Assignment Study Questions for Lesson 33
Exam Final Exam

Assignments:

Unfinished Assignment Study Questions for Lesson 13

Lesson Objectives:

- Express numbers in terms of i
- Find the sum, difference, product and quotient of complex numbers


`i = sqrt(-1)` and `i^2 = -1`.



Express the number in terms of i.

`sqrt(-5)`.

This is the same as `sqrt(-1)*sqrt(5)`. So this is equivalent to `i*sqrt(5)`.



Express the number in terms of i.

`-sqrt(-64)`.

This can be written `-sqrt(-1)*sqrt(64)`. So this is equal to `-i*8` or `-8i`.



Express the number in terms of i.

`sqrt(-72)`.

This is the same as `sqrt(-1)*sqrt(72)` or `i*sqrt(72)`. Since 72 is `8*9` and `9 = 3*3` and `8 = 4*2`, where `4 = 2*2`, then we could pull the 2 and the three out, or pull a 6 out. And we're left with 2 inside.

So, `6i*sqrt(2)`.



In order to find the zeros of functions that are not real consider the complex number system.

The complex numbers are formed by adding real numbers and multiples of i. They're of the form `a+bi`. Where 'a' is the real part and 'b' is the imaginary part. Notice that 'a' or 'b', or both 'a' and 'b', can be zero.



A complex number can be written of the form `a+bi`. They include numbers such as `pi`, `-pi`, `root(5)(3)`, 7.5, -15, `1/2`, `2+i`, `7-i`, `-1/2i`, and `15i`, to name a few.

A real number is a complex number where `b = 0`. So, `pi`, `-pi`, `root(5)(3)`, 7.5, -15, and `1/2` are all examples of real numbers.

An irrational number is not a fraction. So `pi`, `-pi`, `root(5)(3)` can't be written as fractions and they're irrational.

And rational numbers are 7.5, -15, `1/2`, because they can all be written as a fraction.

An imaginary number is a complex number where `b != 0`. So `2+i`, `7-i`, `-1/2i`, `15i` are all examples of imaginary numbers. An imaginary number is called imaginary when `a != 0`. So `2+i` and `7-i` are imaginary, whereas when `a = 0`, it's pure imaginary. `-1/2i` and `15i` are a couple of examples.


Simplify and write your answer in the form `a+bi` where both a and b are real numbers.

So we're given `(-4+3i)+(5+8i)`.

Now we need to start by grouping our real numbers and our imaginary numbers. So we have `-4+5` in one parentheses, and `3i+8i` in another parentheses. And we can add `-4+5` and get 1, and `3i+8i = 11i`. So our answer is `1+11i`.



Simplify and write your answer in the form `a+bi` where a and b are real numbers.

Now we're given `(8+7i)-(6+3i)` and we need to start by distributing the negative sign into our parentheses. So we have `8+7i-6-3i`. And now let's group our real and imaginary numbers. `8-6+7i-3i`. That'll give us `2+4i`.

So our answer is `2+4i`.



If a and b are real numbers, then when we multiply `sqrt(a)*sqrt(b) = sqrt(ab)`.

If a and b are not real, remember to pull out the `sqrt(-1)`, before you multiply.



Simplify. Write your answer in the form `a+bi`. a and b are real numbers.

We're given `sqrt(-9)*sqrt(-81)`.

So if we pull out our `sqrt(-1)`, we have `sqrt(-1)*sqrt(9)*sqrt(-1)*sqrt(81)`. And the `sqrt(-1)` is i, so we have `i*3*i*9`, which is `27i^2`. Now since `i^2=-1`, the answer is `-27`.



Simplify and write your answer in the form `a + bi`, where a and b are real numbers.

`(1+2i)(1-3i)`.

Start by distributing 1 into `1-3i`, and we'll get `1-3i`. And then distribute 2i into `1-3i` and get a positive `2i-6i^2`. And since `i^2 = -1`, we have `-6(-1)` or `1-3i+2i-6(-1)`. Now, `-6(-1) = 6`, and we can add that to the 1 and get 7. And `-3i+2i = -i`, so `7-i` is our solution.


Simplify and write your answer in the form `a + bi`, where a and b are real numbers.

`5+2i^2`.

Remember `(a + b)^2 = a^2+2ab+b^2`. So we can write `5^2+2(5)(2i)+(2i)^2` or `25+20i+4i^2`. And since `i^2 = -1`, we have -4 at the end. So `25+20i-4`, which is the same as `21+20i, which is our solution.



In order to simplify powers of `i`, we need to remember that `i^2 = -1` and that -1 raised to an even power is 1, and -1 raised to an odd power is -1.

So `i = sqrt(-1)` and `i^2 = -1` then `i^3` can be written as `i^2*i`, which is `-1*i` or `-i`.

And `i^4` can be written as `(i^2)^2`, or `(-1)^2`, which is 1.

And then `i^5` can be written `(i^2)^2 * i`, which is `(-1)^2*i`, which is `(1)*i`, which is just `i`.

And then `i^6` can be written `(i^2)^3`, which is `(-1)^3`, which is -1.

And `i^7` is the same as `i^6*i`, which is `(i^2)^3*i`, which is `-1*i`, which is `-i`.

And then `i^8` can be written `(i^2)^4`, which is `(-1)^4`, which is just 1.

And `i^9` can be written `(i^2)^4*i`, which is `(1)*i`, which is `i`. So notice that there's a pattern, `i`, -1, `-i`, 1, and then the pattern starts over again. `i`, -1, `-i`, 1, and then the pattern starts over *again*.


Simplify `i^11`.

`i^11` is the same as `i^10*i`, which is `(i^2)^5*i`, which is `(-1)^5*i`.

And that's the same as `-1*i` or `-i`.



Simplify `i^64`.

This is the same as `(i^2)^32`, which is `(-1)^32`, and since 32 is even, this is the same as positive 1.



Conjugates of Complex Numbers. The conjugate of a complex number `a+bi` is `a-bi`. So `a+bi` and `a-bi` are complex conjugates.

The product of a complex number and its conjugate is a real number.



Simplify and write your answer in the form `a+bi`, where a and b are real numbers.

`(5-2i)(5+2i)`. Remember that `(a+b)(a-b) = a^2-b^2`.

So rather than FOILing, we can take a shortcut and say `5^2-(2i)^2`, which is the same as `25-4i^2`, where `i^2 = -1`, so that becomes a positive. So `25+4` is `29`, which is our solution.



Simplify. Write your answer in the form `a+bi` where a and b are real numbers.

`(7+i)/(-3-4i)`. We can simplify this by multiplying the top and bottom of our fraction by the conjugate of `-3-4i`. That is multiplying the top and bottom of our fraction by `-3+4i`.

So, in our numerator, we'll get `-21+28i-3i+4i^2`, and in the denominator, we'll have the first term squared minus the second term squared.

So this can be simplified to `-21+25i+4(-1)`, and then the denominator, `9-16i^2`, where `i^2=-1`. So we have `9+16` in the denominator, which is 25. And then in the numerator we get `-25+25i`. So 25 can be simplified out of everything, and our final answer is `-1+i`.