### 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`.