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Contents

## All The Numbers

This last lesson before the midterm, sort of stands out on its own, and w're going to do a review section to make sure you've got a grasp on this complex new material that we covered. Just as we started out the course with building up our knowledge of sets of numbers, we're going to round up this section of the class by revisiting this and acknowledging how we've built off of it. As a review of everything that we've learned about sets of numbers, here is a blank version of the full world of numbers diagram. And I would like you to type the name of each set into the proper position on the chart.

## All The Numbers

All of the numbers that we've worked with fall under the category of complex numbers. Complex numbers can have both real and imaginary components. So that means that real numbers and imaginary numbers are each subsets of the complex numbers. Within the real numbers, we have all of the same categories as we did before.

## Everything in its Place

Now that we've filled in the proper sets in the proper places in the world of numbers diagram, let's put some numbers in it. As you did early in the course, please place each of these numbers in the place it belongs on the diagram. Label the diagram with each letter in the appropriate box.

## Everything in its Place

I filled in the numbers in the proper spots. Irrational, integer, whole, natural, rational, complex, and imaginary.

## Complex Number Manipulation

Now that we've looked at complex numbers in our world of numbers diagram, let's manipulate them a bit. Please simplify each of these expressions so that it is written in the standard form for a complex number, a plus bi.

## Complex Number Manipulation

Remember, that when we add and subtract complex numbers, the imaginary parts are like terms and the real parts are like terms. So we end up with 1 plus 3i plus the quanity negative 6 plus 2i equals negative 5 plus 5i. 1 plus 3i minus the quanity negative 6 plus 2i gives us 7 plus i. When we multiply complex numbers, this needs to happen by distributing in the same way that we did with binomials involving, involving variables. The quantity 1 plus 3i times the quantity negative 6 plus 2i, gives us negative 6, plus 2i, minus 18i, plus 6i squared. Which when simplified, gives us negative 6, minus 16i, minus 6 or negative 12 minus 16i. And lastly, when we divided complex numbers, the final thing we need to do is to make the denominator into a real number, so that we can just let it modify the coefficients of the real and imaginary parts of the denominator. To change the denominator in this way, we multiply both the denominator and the numerator by the complex conjugate of the original denominator. So, negative 6 minus 2i in this case. Now let's look at the problem. 1 plus 3i, divided by negative 6 plus 2i, gives us 1 plus 3i, times the conjugate negative 6 minus 2i, divided by negative 6 plus 2i, times negative 6 minus 2i. Simplifying further, we got negative 6 minus 2i minus 18i minus 6i squared, divided by 36 minus 4i squared. Which gives us negative six minus 20i plus 6, divided by 36 plus 4 or negative 20i over 40, and finally negative i over 2.

## Solve For x

Now that we've worked with expressions for a while, let's move on to some equations. Please solve for x in the following equation. 3x squared plus 18 equals 0. Although there are a couple of different methods you could use for solving this, I'd like you to try to do it without using the quadratic formula this time. Please write the solutions in as concise a form as you can.

## Solve For x

So, when we simplify this, we subtract 18 from both sides and we get 3x squared equals negative 18. We then divide by 3 on both sides, we get x squared equals negative 6, giving us our solutions of x equals plus or minus the square root of negative 6. And when we pull out the negative 1 as an i, we get x equals plus or minus i times the square root of 6. This is two solutions, a plus i square root of 6 and a negative i square root of 6.

## Equation of a Parabola

And now for another equation, this time of a parabola. Here is the graph of a parabola for you. Now, first of all, what is the equation of this parabola in vertex form?

## Equation of a Parabola

Your solution is y plus 3 equals negative one third times the quantity x minus 2 squared.

## Roots of the Parabola

What are the roots of this parabola? List them with commas separating them.

## Roots of the Parabola

We can use a quadratic formula to solve this problem. First we put our answer in standard form, plug in 0 for y, and then substitute into the quadratic formula to obtain our two solutions of 2 plus 3i and 2 minus 3i.