Complex numbers are numbers that involve the number i, and where i is the square root of negative 1.
Here are some problems that involve i. First of all, write using i, √(-72).
Now, what this will be, will be, i√(72). So, a negative number under a radical, now,
we take care of that by writing i outside the radical. Sometimes, we write it before the square root; sometimes, we write it after;
it depends on whether it's going to look like i is under the square root or not, because i isn't under the square root.
That's why I write it in front right, here. Now, I'm going to simplify √(72). This will be square root of √(36 x 2)
So, I end up with 6, that's the square root of 36, 6i√(2).
So, this is how I take square roots of negative numbers and write them, and rewrite them using this complex number i.
Now let's take a look at some addition, subtraction, multiplication, and division with complex numbers.
Here, I have written "subtract." I have (6 + 7i) - (4 + i). So, what I'm going, first of all, is take care of this subtraction symbol:
remove these parentheses and change the sign of each thing inside: 6 + 7i - 4 - i. Now, I can combine similar terms
in the same way that I combine similar terms with polynomials: 6 + (-4) = 2; 7i + (-1i) = 6i.
So, this is the complex number I get, 2 + 6i, when I subtract (4 + i) from (6 + 7i).
So, here's the difference of two complex numbers, and it turns out to be a complex number. Here's our next problem:
I want to multiply 6i(4 - 3i). I'm just going to do it by applying the distributive property:
6i(4) = 24i minus, that's that subtraction sign, 6i(3i) = 18i².
Now, i² = -1. So, I can simplify this further, and write it as 24i - (18)(-1).
Now, usually we write the real number part first and then the complex, or the part that involves i second.
So, (-18)(-1) = 18, + 24i. So, this kind of multiplication works just the same as multiplication
with polynomials except that if I run across an i², I replace it with -1,
and that allows me to do a little more simplification. Problem number 4: I have (2 + 5i)².
So, I square the first term, 2² = 4; square the last term, (5i)² = 25i²; and then, I take twice the product of the two terms;
this is (10i)(2) = 20i. So, I square this expression in the same way I would square a binomial;
and now, I have a little more simplifying I can do because i² = -1.
So, I have -25 + 4 = -21, + 20i. So, that's kind of nice, here,
I square a complex number, and I end up with another complex number. So, the only difference between this
and multiplication with polynomials is that every time I see an i², I replace it with -1;
and that gives me a little more simplification to do; -25 + 4 = -21, + 20i.
Here's a division problem: 4/(2 - 3i). We'll accomplish this by a method similar to rationalizing the denominator.
We'll multiply the numerator and denominator by the conjugate of the denominator.
What this does is gets rid of all the i's in the denominator because remember i is a square root;
it's the square root of -1. On top, I have 4(2) = 8, plus 4(3i) = 12i. On the bottom, I have (2 - 3i)(2 + 3i);
so, it's the square of the first term, 4, minus the square of the last term, (3i)² = 9i².
Now, i² = -1, (-1)(-9) = 9. I get (8 + 12i)/(4 + 9); that's (8 + 12i)/13.
So, I'm going to write that over here: (8 + 12i)/13. Now, if I want to write this as a complex number
in standard form, it would be (8/13) + (12/13)i. That's writing it as a complex number in standard form.
This is also the correct answer right here; and I've divided 4 by (2 - 3i), and I end up with this complex number.
The key to division is really in multiplying the numerator and denominator by the conjugate of the denominator.