For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Complex Numbers
 So far, we've found that some polynomials are irreducible in the real numbers: they cannot be broken down into smaller factors and they have no roots. For example, x^{2}+1 has no roots and cannot be broken down any further.
 We introduce a new number to get around this: the imaginary number,
Notice that i^{2} = −1, that is, we can square it and get a negative. That's something we could never do in the real numbers!
√−1
= i.  With this idea in mind, we can now take the square root of any negative number. The negative just pops out as i, while we do the rest of the square root normally. For example, √{−49} = 7i.
 With the idea of the imaginary number i in place, we can create a new set of numbers, which we'll call the complex numbers (denoted by ℂ). We still want to be able to talk about real numbers (ℝ), so we'll need them to appear along with i. Thus we'll make each complex number with a real part and an imaginary part:
where a and b are both real numbers. [a is the real part, bi is the imaginary part.]a + b i,  Working with complex numbers is similar to working with real numbers.
 Two complex numbers are equal when the parts in one number are equal to the parts in the other.
 We can do addition and subtraction by having the real parts add/subtract together and the imaginary parts add/subtract together. Other than the two parts staying separate, it works like normal. Notice how this is similar to adding/subtracting like variables in algebra.
 Multiplication also works very similarly to what we're used to. Just approach it like a FOIL expansion. [Since i^{2} = −1, it will transform during simplification.]
 Division is a little trickier. First, we need the idea of a complex conjugate. The conjugate of (a+bi) is (a−bi) and viceversa. Notice that whenever you multiply a complex number by its conjugate, it always results in a real number. This means if we have a fraction (division) with a complex number in the denominator, we can multiply the numerator and denominator by the complex conjugate for the denominator so that we can now have a real in the denominator. Then we just divide like usual.
 Now that we understand how complex numbers work, we can revisit the quadratic formula and use it to find the roots of "irreducible" quadratics.
Previously, we couldn't use the formula if b^{2} − 4ac < 0, but now we know it just produces an imaginary number! Furthermore, because of the ±√{b^{2}−4ac} part in the quadratic formula, we see that complex roots must come in conjugate pairs. That is, if p(x) is a polynomial:x = −b ±
√b^{2} −4ac
2a
p(a+bi) = 0 ⇔ p(a−bi) = 0.
Complex Numbers

 Standard form is the form a+bi for complex numbers. [In this case, the real part will wind up disappearing (since a=0) because each number is entirely imaginary because of the negative under the square root.]
 For each number, begin by pulling out the negative under the square root as an i.
√−9,
√−32,
√−147⇒ √9 i,
√32i,
√147i  From there, simplify each square root.
[If you're unsure how to simplify square roots, search Educator.com for lessons about it.]√9 i,
√32i,
√147i ⇒ 3i, 4√2 i, 7√3 i

 When adding complex numbers, real parts add together and imaginary parts add together.
 Doing this is as simple as keeping the numbers with an i separate from those without. Think of it as combining like terms: you wouldn't try to combine 1 +2x, so you shouldn't try to combine 1 + 2i.
(3+5i) + (8−7i) = 3+8 +5i − 7i = 11 −2i

 Subtracting complex numbers is very similar to adding them: real parts and imaginary parts stay separate from one another. The only difference is that we need to subtract the complex number instead.
 Don't forget to distribute the subtraction as a negative to both parts of the complex number. You don't want to accidentally only subtract one half of the complex number.
(3+5i)−(8−7i) = (3+5i) + (−8+7i) (3+5i) + (−8+7i) = 3 − 8 + 5i + 7i = −5 + 12i

 When multiplying complex numbers, just expand it like you're used to for polynomials (aka FOIL):
(4−7i)(3+5i) = 12 + 20i − 21i − 35i^{2}  From there, remember that i^{2} = −1 (because i = √{−1}):
12 + 20i − 21i − 35i^{2} = 12 + 20i − 21i + 35  Then finish up by adding together like terms:
12 + 20i − 21i + 35 = 47 − i

 The complex conjugate is the same as the original complex number except the + or − sign on the imaginary portion is flipped to the opposite. For example, the complex conjugate to a+bi is a−bi, and the complex conjugate to a−bi is a+bi. [For notation, occasionally you will see the complex conjugate of a complex number indicated by a bar over it. For example, the complex conjugate of a+bi can be written as ―a+bi = a−bi.]
 To find the conjugates, we just need to flip the sign on the imaginary part.
+5i ⇒ −5i −852i ⇒ +852i  Replace in the original complex number with the flipped version:
(3+5i) ⇒ (3−5i) (187−852i) ⇒ (187 + 852i)

 Remember, we don't intuitively know how to divide by a complex number. However, we do know how to divide by real numbers. Luckily, we can turn the denominator into a real number by multiplying the top and bottom by the denominator's conjugate:
5 3+2i· 3−2i 3−2i  From there, carry out the normal multiplication of complex numbers (but make sure to realize there are implied parentheses for the fraction, so we will use distribution):
5 3+2i· 3−2i 3−2i= 15 + 10i 9 − 6i + 6i − 4i^{2}  Then simplify:
We can't simplify anymore, so we're done now. [It's possible that some teachers or books would require the answer to be precisely in the form a+bi, not allowing fractions over both parts. If that's the case, just put the denominator on each part and get [(15 +10i)/13] = [15/13] + [10/13]i.]15 + 10i 9 + 4= 15 +10i 13

 Just like combining fractions normally, we must have each fraction over a common denominator before we can add them together. This gives us two ways to approach the problem. We could multiply the top and bottom of each fraction by the other fraction's denominator (to guarantee a common denominator), then add them together, then use the complex conjugate to make the denominator a real number. Alternatively, we could make each denominator a real number with complex conjugates, then put them on a common denominator (but a real number), and finally add them together. There's slightly less arithmetic with the second method, so we'll use that instead. Still, the first way would work as well if you wanted to use it.
 Begin by turning each denominator into a real number. Do this by multiplying top and bottom by conjugates:
1+5i 1+5i· 7i 1−5i+ 4−3i −2−i· −2 + i −2 +i7i + 35i^{2} 1 − 25i^{2}+ −8 + 10 i − 3i^{2} 4 − i^{2}−35 + 7i 26+ −5 + 10i 5  Simplify the numerator and denominator where possible:
−35 + 7i 26+ −1 + 2i 1  Put them over a common denominator:
−35 + 7i 26+ −26 + 52i 26  Finally, combine them:
[It's possible that some teachers or books would require the answer to be precisely in the form a+bi, not allowing fractions over both parts. If that's the case, just put the denominator on each part and get [(−61 + 59i)/26] = −[61/26] + [59/26]i.]−61 + 59i 26
 Use the quadratic formula to find the roots:
x = −(−6) ±
√(−6)^{2} − 4 ·1 ·202·1  Work through simplifying it:
6 ±
√36 − 1002= 6 ±
√−642  Before, when we only were allowed to work in the real numbers , we would have said that the polynomial was "irreducible" because of the negative in the root. Now that we're using complex numbers, we see that it's just an imaginary number!
6 ±
√−642= 6 ±8 i 2  Finally, simplify the fraction:
Thus the two roots are x = 3 − 8i, 3+8i.6 ±8 i 2= 3 ±8i
 If a complex number is one of the roots to a polynomial, then that number's conjugate must also be a root. Thus, since x=−3−3i is a root, we know the conjugate x=−3+3i is also a root. [You can see why this must happen from the quadratic formula. If an imaginary number appears in the quadratic formula, it must show up in the square root part. Right in front of the root is the ± symbol, indicating a pair of + and − imaginary parts.]
 Once we know that the two roots are x=−3−3i, −3+3i, we need to verify them. Verify each root by plugging it into the polynomial and making sure it comes out to be 0 (the definition of a root is that when you plug it in, it makes the polynomial 0).
 Let's do x=−3−3i first:
(−3−3i)^{2} + 6(−3−3i) + 18 = (9 +18i + 9i^{2}) −18 − 18i + 18 = (18i) − 18i = 0 \  Next, do x=−3+3i:
(−3+3i)^{2} + 6(−3+3i) + 18 = (9 − 18i +9i^{2}) −18 + 18i + 18 = (−18i) + 18i = 0 \
 When complex numbers are involved, it can be quite difficult to factor in the method we're used to. Luckily, we have another way to find the factors: find the roots first, then use them to get the factors. It's easy to find the roots, just use the quadratic formula:
x = −8 ±
√8^{2} − 4·1 ·182·1  Work through simplifying it:
−8 ±
√64−722= −8 ±
√−82= −8 ±2√2 i 2= −4 ±√2 i  Now we have the roots of the polynomial x = −4 − √2 i and x = −4 + √2 i. However, these are not the factors. Remember from before, if we had some polynomial where t=5 was a root, then (t−5) would be a factor. Thus, we set up (x − root) to find each factor:
Now we have the factors: ( x + 4 + √2 i ) and (x+4 − √2 i).x−[ −4 − √2 i] x− [−4 + √2 i] x + 4 + √2 i x+4 − √2 i  It's always a good idea to check your work when you can, and it's easy to do so here: just multiply the factors together and make sure you get the original polynomial:
( x + 4 + √2 i ) (x+4 − √2 i) x^{2} + 4x − x√2 i + 4x + 16 − 4√2 i + x √2 i + 4 √2 i − 2 i^{2} x^{2} + 8x + 18 \
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Complex Numbers
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro
 Introduction
 A Wacky Idea
 Square Roots and Imaginary Numbers
 Complex Numbers
 Addition and Subtraction
 Multiplication
 Division
 Complex Conjugates
 Division through Complex Conjugates
 Factoring SoCalled 'Irreducible' Quadratics
 But Are the Complex Numbers 'Real'?
 Still, We Won't See Much of C
 Example 1
 Example 2
 Example 3
 Example 4
 Intro 0:00
 Introduction 0:04
 A Wacky Idea 1:02
 The Definition of the Imaginary Number
 How it Helps Solve Equations
 Square Roots and Imaginary Numbers 3:15
 Complex Numbers 5:00
 Real Part and Imaginary Part
 When Two Complex Numbers are Equal
 Addition and Subtraction 6:40
 Deal with Real and Imaginary Parts Separately
 Two Quick Examples
 Multiplication 9:07
 FOIL Expansion
 Note What Happens to the Square of the Imaginary Number
 Two Quick Examples
 Division 11:27
 Complex Conjugates 13:37
 Getting Rid of i
 How to Denote the Conjugate
 Division through Complex Conjugates 16:11
 Multiply by the Conjugate of the Denominator
 Example
 Factoring SoCalled 'Irreducible' Quadratics 19:24
 Revisiting the Quadratic Formula
 Conjugate Pairs
 But Are the Complex Numbers 'Real'? 21:27
 What Makes a Number Legitimate
 Where Complex Numbers are Used
 Still, We Won't See Much of C 29:05
 Example 1 30:30
 Example 2 33:15
 Example 3 38:12
 Example 4 42:07
Precalculus with Limits Online Course
Transcription: Complex Numbers
Hiwelcome back to Educator.com.0000
Today, we are going to talk about complex numbers.0002
At this point, we know a lot about factoring polynomials and finding their roots.0005
Still, there are some polynomials we can't factor.0009
There is no way to reduce them into smaller factors, so they are irreducible; and they simply have no roots.0011
There is just no way to solve something like x^{2} + 1 = 0.0016
There are no roots to x^{2} + 1, because for that to be true, we would need x^{2} = 1.0020
But when you square any number, it becomes a positive; if we square +2, then that becomes +4.0025
But if we square 2, then that is going to become +4, as well; the negatives hit each other, and they cancel out.0033
This pattern is going to happen for any negative number and any positive number, and 0's are just going to stay 0's.0040
So, there is no way we can square a number and have it become negative.0045
But what if that wasn't the whole storywhat if there was some special number we hadn't seen, that, when squared, does not become positive?0050
That is an interesting idea; if that is the case, we had better explorelet's go!0058
We are looking for a way to solve x^{2} = 1; in other words, we are looking for something that is the square root of 1.0063
We are looking for something that, when you square it, gives out 1.0070
So, here is a crazy idea: why don't we just make up a new number?0074
We will try something really crazy, and we will just create a number out of whole cloth.0078
We will imagine a special number that becomes 1 when squared; so we are making a new number.0082
And since we are using our imagination to think of this new thing, we will call it an imaginary number, and we will denote it with the symbol i.0088
We know that i = √1, whatever that means...which means that, when you square i, you get 1.0097
The square root of 4 is 2, because when you square 2, you get 4.0104
So, when you square i (since it is the square root of 1), you get 1.0108
These two ideas are how we are going to define this new thing.0113
When you are writing it, I would also recommend writing it with a little curve on the bottom, just so you don't get it confused.0117
Sometimes, if you are writing quickly, you might end up not even putting a dot,0122
at which point it would be hard to see whether you meant to put down 1 or you meant to put down i.0125
So, I recommend putting a tiny little curve at the bottom, and then a dot like that, when you are writing it by hand.0130
That way, you will have some way of being able to clearly see that you are talking about the imaginary number, and not a normal number.0134
With this new idea of i, we can solve the original equation: since i is equal to √1, we can just take the square root of both sides.0141
And remember: when you take the square root of both sides, you have to introduce a plus/minus.0148
The square root of both sides of an equation...plus/minus shows up, no matter what.0152
So, we get x = ±i; let's check it really quickly.0155
If we have positive i, squared, then that is going to be equal to i^{2}, which is equal to 1, as we just had here.0159
And our other possibility, if we have (i)^{2}...well, negative times negative becomes positive.0168
So, we have positive i squared; but i^{2} is, once again, 1; so it checks outboth of these are, indeed, solutions.0174
So, we have found how to solve x^{2} = 1.0183
We have this new idea of taking a square root of 1, which means, when we square that thing,0186
this thing that we have just created, the imaginary number, we will have 1.0191
We can now use this idea to take the square root of any negative number.0197
We are not just limited to taking the square root of 1; we can do it on anything.0199
We just separate out that √1; that will become an i; and then we take the root as normal.0203
For example, √25 would become...we pull out 1, so that is the same thing as 25 times 1.0207
So, we separate this out; we break it into two square roots (a rule we are allowed to do): times √1.0218
So, √1 becomes i; √25 is 5; so we have gotten 5i out of that.0223
We look at the square root of 98; well, that is the square root of 98, times 1 inside.0230
So, we can separate that out, and we will get √98 times √1.0236
√1 will become i; but what is the square root of 98?0242
I am not quite sure, so we need to break it up a bit more; look at how we can break that into its multiplicative factors.0245
98 is the same thing as 49 times 2; the square root of 49 is 7; 7 times 7 is 49; so we have 7√2i.0250
Finally, the square root of 60; we can see that as 60 times 1, so we separate out the 1 from √60, so this becomes i.0260
What is √60? How can we break that into its factors?0270
Well, we have a 6 times a 10; that is still not quite enough, though, so we break that up some more.0273
We have 2 times 3 for 6, and 2 times 5 for 10, still times i; we see we have a 2 here and a 2 here,0279
so we can pull them out, because they come as a pair.0286
2 times √(3 times 5); we can't do that, so we might as well just turn it into one number, 2√15 times i.0288
So, we can now take the square root of any negative number by having this idea of i, the imaginary number.0295
With this idea of the imaginary number in place, we can create a new set of numbers, which we will call the complex numbers.0302
And we denote it with ℂ; if you are writing that by hand, you make a normal C, and then you make a little vertical line like that.0307
We still want to be able to talk about the real numbers, which, remember, we denote with this weird ℝ symbol.0313
So, we will need them to appear along with i; so we need to have real numbers show up along with this imaginary number.0318
And so, we just saw that we can have imaginary numbers that had this real component multiplied against them.0323
We took the square root of 25, and we got 5i; so we are going to have to have some number times i;0328
and we will also want to have a real part, a; so we will have the real part that will be a.0335
a is the real part; and then we will have the imaginary part that will be bi; bi is the imaginary part.0343
And a and b are both real numbers; they are just coming out of that real set, like usual.0353
This gives us an entirely new form of number: as opposed to just being stuck with real numbers,0358
we have a way of having a real number and a complex number, both interacting with each other.0362
They will come together as a package; and this is the idea of the complex numbers.0366
Two complex numbers are equal when the parts in one number are equal to the parts in the other.0371
If we have a + bi, and we are told that that is equal to c + di, then that means the real parts have to be equal.0375
So, we know that a and c are equal.0383
Also, we know that the imaginary parts have to be equal; both parts have to be equal for this equality to hold.0387
bi and di are the same thing, which means that b and d must be the same thing, since clearly i is going to be the same thing on both sides.0393
Great; all right, so how do we do our basic arithmetic with these things?0399
Addition and subtraction first: we add and subtract complex numbers pretty much like we are used to.0404
a + bi plus c + di just means we are going to combine our real parts (they will become a + c);0408
and our imaginary parts (bi and di) will come together, and we will get (b + d) all times i; bi + di becomes (b + d)i.0415
The same thing over here; if we have a + bi minus c + di, then we have (a  c), and the bi's do the same thing with the di's.0426
So, we are going to have b  d, because remember: there is that minus symbol there; so it is b  d there.0439
a + bi minus c + di...we can also think of that as just distributing this negative sign, like we did before.0445
So, we are now adding a c and a di; that is one way of looking at subtraction.0451
So, real parts add and subtract together, and imaginary parts add and subtract together.0456
Other than the fact that they stay separate, it pretty much works like normal.0460
So, as long as they stay separatethey stay on their two sidesimaginaries can't interact directly with the reals;0464
reals can't interact directly with the imaginaries when we are keeping it in addition and subtractionit is pretty much just normal.0469
Let's look at two examples: 5 + 2i plus 8  4iour 5 and 8 will be able to interact together, because they are both real numbers.0475
And let's colorcode that, just so we can see exactly what is going on.0483
So, with our colors before, with red representing reals again, we have 5 + 8, and then it will be +, and then our imaginaries interact as well; 2i  4i.0486
5 + 8...we get 13; and 2i  4i (because this was a 4i here)...we get 2i, which we could also write as 13  2i.0497
13 + 2i and 13  2i mean the same thing; great.0511
If we did it with subtraction, (5 + 2i)  (8  4i), then we can distribute this negative sign; and we get 8,0516
and then we will cancel out the minus there, and we get + 4i; so 5  8 + 2i + 4i.0524
5  8 becomes 3; plus 6i; great.0540
All right, multiplication is pretty much going to work very similarly, as well.0547
Now, notice: we have two things in this sort of factorlooking form; so we do it as a FOIL expansion.0551
We are going to do that same idea of distribution.0558
We will multiply again in the same way that we distributed before.0559
a will multiply on c, and a will multiply on di; so we get ac and adi.0563
And then, bi will multiply on c, and bi will multiply on di; so we will get...bi on c gets us bci, and bi on di will get us bdi^{2}.0570
Now, remember: i^{2} = 1; so when we have this i^{2} here, it cancels out, and it is like we have subtraction.0581
So, we put together our real components now: bd and ac gets us ac  bd, because it was bd.0590
And we have our imaginary components, adi and bci, so we have (ad + bc)i.0599
Remember, this i^{2} equals 1, so it will just transform during the process of simplification.0606
Now, you could memorize this formula right here, but I wouldn't recommend memorizing this formula right here,0611
because you already know the FOIL expansion, and as long as you can remember i^{2} = 1, that will keep it easier.0616
That is the better way to do it.0621
All right, let's see some examples: (1 + 2i)(5  i): 1 times 5 becomes 5; 1 times i becomes i;0623
2i times 5 becomes 10i; 2i times i becomes 2...and we have two i's, so 2i^{2}.0632
Once again, i^{2} is equal to 1; that cancels out and becomes a plus.0639
So, 5  i + 10i + 2: 5 and 2 combine to give us 7; and i + 10i combine to give us + 9i.0645
Great; the next one is (6 + 10i)(5 + 3i): 6 times 5 is 30; 6 times 3i is 18i; 10i times 5 is 50i; and 10i times 3i is 30i^{2}.0658
Once again, remember: i^{2} is 1, so we have 30, at which point our 30 and positive 30 cancel each other out.0674
And we are left with just 18i + 50i, so we get a total of 68i.0681
Great; the final one is division; now, division is a little more tricky.0686
Consider if we had (10  15i)/(1 + 2i): now, at first, we might think,0690
"Oh, we have 10 and 1; we have 15 and 2i; so we will get 10/1 and 15/2,0695
because the i's will cancel out"; but that would be wrong.0702
We have to divide by the entire denominator, not just bits and pieces.0704
For example, to see why this has to be the case, imagine if we had 5 + 5, over 3 + 2; that is really 10/5, which equals 2.0708
But we could get confused and think that that was going to be 5/3 + 5/2; but division does not distribute like that.0719
We are not allowed to do that; so we can't do the same thing here with our 1 + 2i.0729
We can't break it up and distribute the pieces, because it is nonsense in real numbers; so it is definitely going to be nonsense in the complex.0733
What if we break it up, and we put 1 + 2i onto the 10, and then we put 1 + 2i on the 15i separately?0740
We get 10/(1 + 2i), and we get + 15/(1 + 2i).0747
Well, that is true; we broke it up; we can do that with normal things.0754
We could have it, if we wanted to, going back to 5 + 5 over 3 + 2we could have that as 5/(3 + 2) + 5/(3 + 2).0758
But that doesn't help us; we still have to divide, ultimately, by this 1 + 2i.0766
We don't know how to divide by a complex number yet; that is our problem.0774
Simply put, we have no idea how to divide by complex numbers; that is our problem.0779
Addition and subtraction made natural sense; real numbers stuck together; imaginary numbers stuck together.0786
FOIL was able to allow us to do multiplicationwe just did normal distribution, and we remembered the rule that i^{2} becomes 1.0791
But division...we don't have a good understanding of what it means to divide by a complex number.0797
That is tough; now, what we could do, if there was some clever way to get rid of having a complex number in the denominator0803
if we could somehow make it into an alternate form where we disappeared the complex number in the denominator we would be good.0810
Hmm...to figure out this clever method that we want, first notice something you might have seen while we were working on quadratics.0818
If you have (x  2)(x + 2), you get x^{2}, and then + 2x, but also  2x;0824
since we have the 2 and the +2 here, they end up canceling each other out, and so we are left with just x^{2}  4.0830
And there is no middle term with just x; there is no x that shows up.0837
We are able to get rid of it, and have only the doubled and then no x whatsoever.0843
We can expand this idea to complex numbers: we do a similar pattern, (1  2i)(1 + 2i); let's work that out.0848
We get 1  2i + 2i  4i^{2}; so +2i and 2i cancel each other out, because we have the negative here and the plus here.0855
And then, we have 4i^{2}, so that will cancel and become a plus; and we get 1 + 4 = 5.0864
So, we have been able to figure out a way to multiply this thing and get just a real number.0870
So, if we use this pattern, (a + bi)(a  bi), you automatically get a real number.0875
When you multiply it out, it results in a number that has no imaginary part.0881
You can get that i to disappear entirely, and get something that is completely real.0884
This idea we call the complex conjugate; it comes up often enough, and it becomes important enough,0890
that we give it a special name (complex conjugate); and we also give it a special symbol.0895
We denote it with a bar over the number; so if we have a + bi as our complex number,0899
we can talk about its conjugate with a bar over it; and that is a  bi.0904
The conjugate of a + bi is a  bi; and what is the conjugate of a  bi?0909
Well, we just flip it again, back to a plus, vice versa; a + bi is the conjugate of a  bi; the conjugate of a  bi is a + bi.0913
They just end up flipping between each other, as long as we are doing conjugates.0920
Notice that, whenever you multiply a complex number by its conjugate, it always results in a real number.0924
So, by multiplying with the conjugate, we can get rid of imaginary things; we can get rid of it.0930
Multiply by a conjugate; you always get a real number out of it.0937
So, let's look at that: a + bi times a  bi: we get the a^{2}, and we will get  abi + abi;0941
plus here, minus here; those cancel out; i^{2} is a 1, so that causes that to become a plus.0948
So, we will end up with a^{2} + b^{2}, and no i; a and b were just real numbers, so we have something that is entirely real.0955
We started with imaginary things, but by multiplying these together,0963
choosing carefully what we had, we were able to knock out imaginaries entirely and get something that is just real.0966
With this idea of complex conjugates in mind, we can now deal with division.0972
We simply turn the denominator into a real number by multiplying top and bottom of the denominator's conjugate.0976
We want to get the bottom to turn into just a real number, because we know how to divide by reals; you just put it on a fraction.0981
So, with c + di, we need to multiply c  di.0987
Of course, we can't just multiply the bottom because we feel like it;0991
so we also have to multiply the top by c  di, as well, because something over itself is always 10994
(as long as you didn't start with 0/0; then the world explodes).1000
But as long as it is something over something, and that something isn't 0, you get 1.1003
So, c  di over c  di...we can do that; we just trust, intrinsically, that dividing something by itself is 1.1006
That is the nature of division; that is the point of it.1012
So, ac + bd + bc  ad times i over c^{2} + d^{2}; that is what it will end up simplifying to.1015
And we could work this out, and we would see that this formula ends up working out.1023
I don't want you to memorize this formula; I don't even really see a good point to working through it.1027
The important thing to know is: just remember to multiply top and bottom.1030
Remember to multiply top and bottom by the denominator's conjugate.1034
This idea of being able to multiply by a conjugatethat is the really cool thing.1039
You could memorize a formula, but it is not going to help you to memorize a formula, because it is hard to recall a formula like this.1043
It is much easier to remember that I have division of a complex number.1050
I multiply by the conjugate, because I want to get rid of real numbers.1054
I have to multiply top and bottom, though, because of course, if you did otherwise, you would just be playing fantasy.1057
You have to multiply by the same thing on the top and the bottom to keep it what you started with.1061
All right, let's see an example: (10  59) over (1 + 2i).1064
We want to multiply by the conjugate of (1 + 2i) (if we wanted to, we could express that with a bar all over the top of it), which would be (1  2i).1070
We multiply by that, and we know we will have gotten to just a real number.1079
1  2i multiplies top and bottom; and it does have to come in parentheses, because it is a whole thing multiplying some other whole thing.1083
You don't just get multiplied bits and pieces.1090
We work that out: 10 times 1 gets us 10; 10 times 2i gets us 20i; 15i times 1 gets us 15i; 15i times 2i gets us +30i^{2}.1093
What is on the bottom? We have (1 + 2i)(1  2i); 1 times 1 gets us 1; + 2i,  2i; those will cancel out; 2i + 2i, and then 4i^{2}.1107
Remember, i^{2} becomes 1; so we cancel out like that.1126
And then, we also see that 2i + 2i cancel each other out.1132
What does this become next? We combine things: 10  30, our real parts on the top, become 20.1135
20i  15i becomes 35i; what is on the bottom? 1 + 4 is 5, so we can divide 20/5, minus 35i/5; so that gets us 4  7i.1141
Great; all right, now that we understand the basics of how to work with complex numbers,1162
we are now at a point where we can actually see how to factor irreducible quadratics.1168
It is now possible for us to factor previously irreducible quadratics and find their roots.1172
So, x^{2} + 1, we see, is now factored into (x + i) and (x  i).1176
Let's check this: we get that this would be equal to x^{2}  ix + ix  i^{2}.1180
Oops, ix; I didn't write that whole thing.1189
Those cancel each other out; i^{2} becomes + 1, so we get x^{2} + 1.1193
Sure enough, it checks out; and we have found a way to be able to factor this thing that, before, we could not factor.1201
It used to be irreducible, but now we see that, through the complex numbers, it is not irreducible at all.1206
It is totally factorable; we can revisit the quadratic formula and use it to find the roots of these supposedly irreducible quadratics.1210
What used to be irreducible for us is no longer, so we can use the quadratic formula.1216
Previously, we couldn't use it when b^{2}  4a was less than 0, because there was no square root of a negative number.1221
But now we know that that just means an imaginary number; so if our discriminant, b^{2}  4a, shows it is less than 0,1226
then that means, not that we have no answers, but that we just have imaginary answers.1232
Cool; furthermore, because of the ± √(b^{2}  4ac) part in the quadratic formula,1236
we see that complex conjugates must come in conjugate pairs.1242
If b^{2}  4ac was less than 0, so this gives out stuff times i, then we have this ± thing;1246
so it is going to be plus stuff(i), minus stuff(i); so we have one version that is a +i and one version that is a i.1254
That is what happens when we are doing a conjugate pair.1260
We have a + bi; its conjugate is a  bi, so if we have stuff + stuff(i) and minus stuff(i), that is what we have right there.1264
All right, so if we have a polynomial where we know that a + bi is a root1274
(that is to say, when you plug it in you get 0), then we know that a  bi has to also be a root;1279
these things come in conjugate pairs all the time.1285
So, we talked a lot about the complex numbers; but we probably have this nagging question in the back of our head.1288
Are they real? They are clearly not real numbers, because we are saying that they are not the real numbers,1294
which are numbers like 5, 0, π, √2...we have been working with them all up until now.1299
But are they realare they legitimateare they something that we really can use,1306
and not be thinking that we shouldn't be using these?1310
I mean, they have the word "imaginary" in their definition; do we really want to be trying to do science or math with something that is inherently imaginary?1313
Let's think about this: what does it mean for a number to be legitimate?1323
What is this idea of a number being a legitimate number that is valid for science, valid for math?1327
Now, we probably all agree that 1, 2, 3...those are totally valid.1333
You could pick up one rock; you could pick up two rocks; you could pick up three rocks.1337
We could actually have these things in our hands and say, "Look, I have that many objects."1341
And we might not be able to pick up 5 billion rocks, but we can get this idea that we could count that many rocks in front of us.1346
So, that seems pretty valid; these nice whole numbers are perfectly reasonable.1352
But what about 1/2 being a real number? 1/2 seems pretty valid, because we could take a pizza, and we could cut the pizza in half.1357
We take a pizza; we cut it down the middle; and now, all of a sudden, we have two chunks of pizza.1366
1 here; 1 here; we are left with two objects that come together to form a whole.1372
But at the same time, we could say, "Well, this is one object, and this is one object; so it is 1 and 1."1378
But we could also say it is 1/2 of what we originally started with, and it is 1/2 of what we originally started with.1384
So, it is 1/2 and 1/2; so it is a little bit more questionable that this is valid,1390
because can you actually pick up a halfobject? No, it is an object in and of itself.1397
But it is connected to other things, so it is not perfectly validnot as valid as the rocks.1401
We can grab rocks; we can hold rocks; but we can definitely believe in halfnumbers.1405
We can believe in rational numbers; we can believe in fractions; it seems reasonable.1410
Well, OK, what about something even more slipperywhat about the negative numbers?1414
Negatives are pretty bad; or we could go even worse, and we could talk about irrational numbers.1419
How can you possibly hold 1 rock? What does that mean?1425
Can you hold √2 rock? Can you hold π rock?1428
You can't hold these things in your hand; so are they valid?1431
We can't cut a π slice of pizza; we can't cut a √2 slice of pizza; what does this mean now?1433
Are they really valid? We certainly used them a lot beforewe are used to using them.1440
So, they seem reasonable in that way; but are they things that are real in the real world?1445
Don't worry; they are not illegitimate; it doesn't mean that they are illegitimate because you can't hold them in your hand.1451
We can use them to represent things in the real world.1456
We can talk about an object falling with negative numbers; we can say it is going a negative height.1458
As opposed to a positive height, where it goes up, it goes a negative height, where it goes down.1464
Or maybe you have $100 in your bank account, but then you pull out 150; that leaves you with negative $50 in your bank account.1468
So, you have an overdrawn bank account; we talk about that with negative numbers.1474
So, that seems pretty reasonable; we could also talk about √2.1477
√2 is able to connect the sides of a square.1481
If we have a square where all of the sides are the same on our square, then the connection between one side and the diagonal is side times √2.1484
So, we can figure that out from the Pythagorean theorem.1498
That makes sense; there is some stuff going on.1500
Or if we go ahead and we look at a circle, we will see π showing up.1502
If we want to talk about the circumference of a circle (pardon my circle; it is not quite perfect), it will be π times 2 time the radius.1506
So, there is π showing up; or, if we wanted to talk about the area, it will be π times the radius squared.1515
So, there are relationships going on in circles.1520
And circles are reallife things; we see circles in lots of places; we see spheres and other circular objects in lots of places in real life.1522
So, it seems reasonable to count √2, π, 1...they are all valid numbers,1529
not because we can hold it in our hand, but because we can use it for totally reasonable things.1534
So, ultimately, these numbers are "real" (not to say real numbers, but "real" numbers, numbers that we believe in),1539
because they have meaningbecause they are useful for something.1545
A number is valid, not because we can hold it in our hands, but because it is useful and/or interesting.1550
That is what makes a number a valid number that we want to work withbecause we can either use it in real things,1557
or it is really interesting and fascinatingit is telling us cool stuff.1562
After all, math is a language; and in language, we can talk about things that aren't just concrete.1565
You can talk about things like "cat" and "tree"; but at the same time, you can also express abstract conceptsthings like "justice" and "freedom."1571
You can walk down the street, and you can point at a cat, and you can point at a tree.1581
But you can't really hold a justice in your hand; and you can't say, "Oh, look, here is a freedom."1585
They are not things that you can hold; they are not tangible, real things.1590
They are abstract concepts that require us to think in this other way.1594
And that is how the numbers work: 1, 2, 3...they are representing concrete things that we can really hold.1598
But we can also talk about abstract ideas, like √2 or π, that are telling us relationships that are really useful.1603
We might not be able to hold it in our hand; but it is still a really useful idea.1609
So, it is just as valid; "cat" and "justice" are both valid things, because they are useful to us.1613
They represent something worthwhile; they represent something interesting.1619
It is the exact same way with the complex numbers; this is how it is with the complex numbers.1624
You can't hold i rocks in your hands; you can't hold 52i in your bank account.1628
But they still have validity; there is still meaning there; they are still valid; they still have meaning.1634
In fact, they have direct connections to the real world; so that might be our other issue:1641
"OK, I can believe in the fact that numbers get to be valid when they are interesting; but are they usefulcan we use them in the real world?"1646
Sure enough, you can: complex numbers show up a lot in electrical engineering.1653
They show up in advanced physics; and they show up in other fields of science.1657
They also show up in lots of advanced mathematics.1660
If you are interested in mathematicsin the really, really high, interesting stuffcomplex start to show up a lot.1663
They are totally valid; you can prove real things; they are really meaningful.1667
By using complex numbers, we can actually model realworld phenomena; and we can make accurate predictions.1671
Complex numbers are proven to be useful; we can actually use a complex number and get truth out of it that we can then measure in the real world.1677
You don't get a complex number of things; but you can have a complex number help you on your way1685
to finding an accurate measurement, to finding something and predicting something that actually works.1690
So, complex numbers are totally valid in terms of being useful in the real world, and also just as a thought construct.1694
In many ways, the name "imaginary" is unfortunate: they are not imaginary in terms of "they don't count; they aren't really there."1700
They are just imaginary because the name stuck; there is no reason that they are less valid than real numbers.1709
They aren't less valid; they are just as valid as any other number.1714
They are not real numbers, which is to say they are not ℝ; they are not those numbers that we talked about before.1717
But the complex numbers can still represent reality.1723
So, they are not real numbers, but they still show reality.1725
They are imaginary, but only in name; they are actually things that can be used to show real life.1728
They tell us all sorts of useful things, and they are pretty cool.1734
Complex numbers are legitimate and valid; they are not real numbers, but they are "real" in the sense that they are a part of the real world.1737
All that said, nonetheless, complex numbers are not going to be something that we will see a lot.1746
They are totally legitimate; they are valid; but we won't see much of them.1751
Complex numbers tend to be connected to advanced math, for the most part.1755
And so, it is really going to be more advanced math than we want to study right now.1758
So, if you keep going in math, or you keep going and see some really highlevel science at some point in a few years,1763
you will probably end up seeing complex numbers be used for real things.1768
But right now, we are just sort of saying, "Oh, lookcomplex numbers! That is cool," and we are moving on to something else.1772
So, most math coursesespecially courses at this levelwill limit themselves to just the real numbers,1778
because if they go too far, it will get too complex (get the joke?).1783
Unless a question specifically asks about complex numbers, or they were directly mentioned in the lesson1791
(such as this one), just stick to the real numbers.1796
You really want to just stick to the real numbers, unless you are working specifically with the complex numbers,1798
or you have been told to work specifically with the complex numbers.1803
We will briefly play with complex numbers in a couple of lessons in this course.1805
But they are something best explored later on in a more advanced mathematics course, or an advanced science course.1808
Thus, in general, limit yourself to using just the real numbers, ℝ, for now.1814
And really, that is going to be pretty easy, because it is what you are used to doing.1819
You are used to just working with the real numbers; so it is not going to be hard to just go back to working with the real numbers,1823
because it is what you have been doing for years and years and years.1827
All right, we are ready for some examples.1830
Simplify (25  45i)/(3 + 4i); remember, we need to multiply by the conjugate.1832
The conjugate to 3 + 4i, which we could denote with a bar over all the top of it, is equal to 3...1838
and then we flip the sign on the imaginary part, so it will be  4i.1844
So, we want to multiply this by (3  4i)/(3  4i).1848
Now, notice: you have to put parentheses around all of this, because the whole thing is multiplyingnot just bits and pieces, but the whole thing.1856
So, we work this out; 25 times 3 becomes 75; 25 times 4i becomes 100i; 45i times 3 will become positive 45...1864
that is 3 times 5 off of 150, or + 135i; 45 times 4 becomes positive, so we will get 4 times 5 off of 200, so 180i.1878
Divided by...3 times 3 gets us positive 9; 3 times 4i gets us + 12i; +4i times 3 gets us 12i; 4i times 4i gets us 4i^{2}.1896
So, we see that we have 12i + 12i, and also when we have i^{2}, it becomes positive.1911
Oops, I accidentally made a typo here: 45 times 4i will become + 180i^{2}.1916
So, cancel out that i^{2}; we get 180; now, let's combine things.1923
75  180; that will get us 255; 100i + 35i will get us +35i; what is on the bottom?1928
4^{2}...we missed that; sorryone more mistake; 4 times 4 gets us 4^{2}i^{2},1938
so 4^{2} gets us 16; 9 + 16 is in our division, so divide by 25.1949
255 + 35i; divide by 25; we notice that we can pull out a 5 from all of these; this is 5 times 51.1957
This is 5 times 7; this is 5 times 5; so we go through and cancel one of the 5's on all of them.1964
And we are left with 51 + 7i, all over 5, which, if we wanted to, we could alternately represent as 51/5 + 7/5 i,1972
keeping our imaginary part and our real part completely separate.1987
Both of these are totally legitimate answers; we would know what we were talking about in either case.1990
All right, the second example: Given that x = 2 + i is a root to the below polynomial, find the other root and verify both.1994
Remember: if x = 2 + i is one of our roots, the conjugate is also the case.2002
So, x bar, the conjugate of x being 2 + i, is going to be...what is the conjugate of that?...2  i.2007
So, we know what the other root is; the other root is 2  i, and our first root is 2 + i.2015
We are guaranteed that a complex conjugate must be the other root, from what we talked about earlier.2021
So now we are told to verify both of them.2027
There are two different ways we can verify this.2029
First, we could verify this through factors; we could show that, if we were to use these as factors...2031
because remember, knowing a root tells you a factor; remember, if we know that there is a root at k,2038
then we know that there is a factor, (x  k); so if we know that there is a root at (2 + i),2046
then we know that there is a factor of (x  2 + i), following that same pattern of x  k.2051
It is just that k, in this case, is two things.2058
That is times (x  (2  i)) for our other factor.2061
So, if we can multiply these two factors together, and we can get x^{2} + 4x + 5,2067
then we will have verified that those must be the roots, because they are the factors,2071
and there is this deep connection between roots and factors; you can go either way.2075
So, let's work this out: simplify the insides first: x minus a negative will become + 2  i;2078
times x minus a minus will become + 2 + i; we can start working this out.2086
x times x becomes x^{2}; x times 2 becomes + 2x; x times i will become + ix.2093
2 times x will become + 2x; 2 times 2 will become + 4; 2 times i will become + 2i; i times x will become ix;2100
i times 2 will become 2i; i times +i will become i^{2}.2109
i^{2} becomes +1, because the i^{2} cancels out.2116
And now, let's work through and see this.2120
So, let's simplify this: x^{2}: how many other x^{2}'s do we have?2122
That is the only one, so we get x^{2} + 2x; how many other x's do we have?2125
We have x there, 2x there, and no other x's; so we put those all together, and we get + 4x.2130
ix'show many ix's do we have? We have that ix and that ix, so ix  ix.2137
They cancel each other out, and they completely nullify each other; so we don't have to put them down at all.2142
How many constants do we have? 4 there; don't forget the 1 that came out of our i^{2}.2147
So, we have 4 + 1, because it flipped the sign; that is + 5.2152
And then 2i  2i; once again, they nullify each other, so we get x^{2} + 4x + 5; it checks out; great.2156
We found the answer.2163
The alternate way that we could do this is: we could do this by verifying that they are, indeed, roots.2164
So, we could do this another way by showing that they are roots; let's start by showing that x = 2 + i is a root.2171
We plug that in; x^{2}, (2 + i)^{2}, plus 4(2 + i), plus 5.2182
(2 + i)^{2} becomes: 2 times 2 becomes positive 4; 2 on i, plus i on 2, becomes 4i; i on i becomes + i^{2}.2192
And then, continue on: plus 4 on 2 becomes + 8; 4 on i becomes +4i; and pull down the 5.2204
i^{2} becomes 1; notice that we have + 4i  4i, so they eliminate each other here and here.2213
4  1 becomes 3; 8 + 5 becomes 3; and we get 0...sure enough, that is a root, because it produces 0.2223
The other one: let's plug in x = 2  i; we plug that one in: (2  i)^{2} + 4(2  i) + 5.2232
2 times 2 is positive 4; 2 on i and i on 2 get us + 4i; i on i gets us + i^{2}.2244
Plus 8, minus 4i, plus 5...so we see that we have a positive 4i here and a negative 4i here; they eliminate each other.2253
We have this i^{2}; it becomes 1; so 4 and 1 gets us 3; 8 and 5 gets us 3, which, once again, equals 0; so they are both roots.2264
There are two different ways to do it: we can show that these are the factors that would be given by those roots,2276
and when you multiply those factors, you get back exactly to where you started; that checks out.2280
Or alternately, we can do it by roots and show that when you plug that in, you get the zeroes; so that checks out.2286
Great; the third example: Factor x^{2}  8x + 19.2291
Well, we know that this is probably going to involve complex numbers; it is probably a little bit hard to figure it out in terms of complex numbers.2297
But can we find the roots? Sure enough, we can find the roots.2302
Let's find roots, and then we will use the roots to give us factors.2305
Remember: once you know roots, you know factors; so we find the roots first.2309
We can just use the quadratic formula, because now we can use it on anything.2313
We don't have to worry about if it is a complex or not.2317
The discriminant won't hold us back, because now we can just get imaginary answers, as well.2319
We have x =...the roots occur at [b ± √(b^{2}  4ac)]/2a.2324
And hopefully, you were able to say that out loud before I said it, to yourself, because really, you want to have that one memorized.2335
I said it the last time we talked about the quadratic formula.2342
The quadratic formula comes up enough in math and science that it is ultimately something you really want to have memorized.2344
All right, so what is our b? Our b is 8.2349
So, we plug that in: [(8) ± √((8)^{2}  4 (what is our a? our a is a 1) (1) (times...what is our c? c is 19)(19)...2353
let's move that square root over all of the way; 2 times...a is 1 again, so 2 times 1.2367
That equals (8) (gets us positive 8), plus or minus the square root of...64; what is 4 times 19? that is 76, so minus 76; all over 2.2373
We divide out the 2, so we will get 8/2; that gets us 4; plus or minus the square root of 64  76; that will still be over 2.2387
Let's put it over that, just so we don't forget that.2395
64  76 gets us 12; so we have 4 ± √12...so we can pull that out as an i, so we will get √12 i, over 12,2397
equals 4 ± √12...what is √12? √12 we can see as √4(3), which equals 2√3,2412
so plus or minus 2√3 i, over 2; look, we have 2 and 2; those cancel out, and we are left with all of our roots.2425
They are when x is equal to 4, plus or minus the square root of 3, times i.2436
Those are our roots; however, those aren't our factors.2442
We want to find what the factors are; so let's get that in another color.2446
If we know that our roots are 4 ± √3i, remember: if you know k is a root, then that tells you x  k is a factor.2450
So, in this case, our roots are x = 4 + √3i, and x = 4  √3i, which is good, because they came as a conjugate pairing there.2463
So, those are both of our possibilities; those are both of our factors.2477
x  k: our factors will be x minus this one right here, so minus (4 + √3i)...not that whole thing...2480
I put that parenthesis on the wrong place; i...the parentheses close there; times (x  this thing here, (4  √3i).2494
So now, let's simplify it, so we can get the factors in a nice, slightlysimpler form to look at.2506
x  4  √3i and x  4 + √3i; we have factored it by being able to do that.2510
And if we wanted to, we could also expand this and check this.2522
And we would be able to show that that is, indeed, exactly what it is; great.2524
The final example: What is i^{3}, i^{4}, i^{5}, i^{6}, i^{7}, i^{8}, etc.?2528
What pattern appears as we go through these powers of i?2535
Let's take a look at how we work through it.2538
If we have i^{1}, just plain i, we have i.2541
That is just what it is; it is just i.2545
What about when we have i^{2}? Well, by definition, that was 1.2547
So, let's see the way it keeps going as we take this up.2551
i^{3}...we multiply the 1 by one more i, so we would get 1 times i, or just i.2553
i^{4} would be equal to...i times i gets us i^{2}; i^{2}(i^{2}) cancels, and we get positive 1.2559
i^{2} cancels, and we get positive 1; so we are left at 1, just a plain + 1.2567
What if we keep going? i^{5} is equal to...well, we multiply by 1, so it is just i, once again.2578
i^{6} would be equal to i^{2}, multiplying by one more i, which we know is 1.2584
i^{7} is equal to i^{3}, which is equal to...we already figured this out; that was i.2590
i^{8}...well, that is going to be equal to i^{4}, because we just multiply the one above.2597
We already figured out what i^{4} is; that is going to be positive 1.2602
Let me make that plus sign a little clearer.2607
i^{9}...if we just kept going, we would have i^{5};2608
we already figured out what i^{5} wasthat was i^{1}, which is just i; and so on, and so on, and so on.2611
So, the pattern repeats every 4.2617
What we need to do is: we basically need to divide by 4 and see what we have.2627
What we can do is divide the exponent of i by 4; then, what do we do next?2632
Let's do a quick check: if we did i^{9}, 4 goes into 9 how many times?2643
It goes in twice; so we would have 8; 9  8 is 1, so we would get a remainder of 1.2648
So then, you look at the remainder, and that tells you that it is equal to i to whateveryoujustfiguredoutyourremainderis.2653
So, for example, if we wanted to figure out what i^{80} is (which is divisible by 4),2674
we can see that is just i to the 4 times 4 times 4 times 4 times 4; if we figure that out for i^{80},2681
then we can figure out that what that is equivalent to...by 4...how many times does that go into 80?2687
4 goes into 8 twice, so that gets us 8  0; bring down the 0; 0; we get 20, and our remainder is 0.2692
So, that would be the equivalent of i^{0}, which is just the same thing as i^{4}, which is +1.2700
So, that is how you want to do it if you are given a really, really, really large i.2708
It is just a question of if you divided it by 4what would be left over? What would be the remainder?2712
And if you end up having a remainder of 0, then it fit perfectly, so it ends up coming out just as 1.2716
All right, great; we will see you at Educator.com later.2721
And we will finally see how complex numbers tell us something about polynomials, more than just quadratics.2723
We will see how they are deeply connected to everything that we have been talking about.2728
It will be so deep that it is called the fundamental theorem of algebra.2731
All right, see you latergoodbye!2735
2 answers
Last reply by: Tiffany Warner
Thu Jun 2, 2016 6:39 PM
Post by Tiffany Warner on June 1, 2016
Hi Professor!
In one of the practice problems, we are asked to find the roots of x^26x+20
The answer I continuously got was 3+sqrt(11)i and 3sqrt(11)i
The answer given does not match. I looked at the steps and there’s two places where I think I’m missing something. Under the square root they show 36100 but I got 3680. I’m not sure where the 100 is coming from.
Then in the last step I noticed the first real number is divided by the denominator, but the real number in the “imaginary part” was left alone. Can we do that?
I often make small silly mistakes that lead me to strange conclusions so I’d thought it best to get clarification.
Thank you very much for your time and help!
3 answers
Last reply by: Professor SelhorstJones
Mon Feb 16, 2015 12:14 PM
Post by Andre Strohhofer on June 19, 2013
How would you graph these numbers, or how would the roots show up on a graph?