For more information, please see full course syllabus of Trigonometry
For more information, please see full course syllabus of Trigonometry
Complex Numbers
Main definition and formulas:
 Complex numbers are numbers of the form x + yi, where x and y are real numbers, and i satisfies i^{2} = − 1.
 The powers of i satisfy the following pattern:
i^{0} = 1 i^{1} = i i^{2} = − 1 i^{3} = −i i^{4} = 1 i^{5} = i i^{6} = − 1 i^{7} = −i i^{8} = 1 :  Operations:
Addition: (x + yi) + (a+bi) = (x +a) + (y + b)i Subtraction: (x + yi) − (a+bi) = (x −a) + (y −b)i Multiplication: (x + yi)(a+bi) = (xa −yb) + (xb + ya)i Conjugation: x + yi= x −yi x −yi= x + yi (x+yi)(x−yi) = x^{2} + y^{2}  Division:
x+yi a+bi= x+yi a+bi· a+bia+bi= x+yi a+bi· a−bi a−bi= (xa +yb) + (ya −xb)i a^{2} +b^{2}= ( xa +yb a^{2} +b^{2}) + ( ya −xb a^{2} +b^{2}) i
Example 1:
Solve the following equation for the complex number z:

Example 2:
Expand and simplify (1+2i)^{3}.Example 3:
Simplify the following powers of i: i^{− 6}, i^{19}, i^{33}, i^{− 13}Example 4:
Simplify the following expression:

Example 5:
Find all complex numbers z satisfying z^{2} = 3 + 4i.Give me an extra page here.
Complex Numbers
 Subtract 3 − 2i from both sides
 (1 − 2i)z = (2 − 8i) − (3 − 2i) ⇒ (1 − 2i)z = − 1 − 6i
 Divide both sides by 1 − 2i
 [(1 − 2i)/(1 − 2i)]z = [( − 1 − 6i)/(1 − 2i)] ⇒ z = [( − 1 − 6i)/(1 − 2i)]
 Multiply by the conjugate ―1 − 2i
 z = [( − 1 − 6i)/(1 − 2i)] ·[(―1 − 2i )/(―1 − 2i )] ⇒ z = [( − 1 − 6i)/(1 − 2i)] ·[(1 + 2i)/(1 + 2i)]
 z = [( − 1 − 2i − 6i − 12i^{2})/(1 + 4)] = [( − 1 − 8i + 12)/5]
 Add 4 + 3i from both sides
 (1 + 3i)z = (3 + 5i) + (4 + 3i) ⇒ (1 + 3i)z = 7 + 8i
 Divide both sides by 1 + 3i
 [(1 + 3i )/(1 + 3i )]z = [(7 + 8i)/(1 + 3i )] ⇒ z = [(7 + 8i)/(1 + 3i )]
 Multiply by the conjugate ―1 + 3i
 z = [(7 + 8i)/(1 + 3i )] ·[(―1 + 3i )/(―1 + 3i )] ⇒ z = [(7 + 8i)/(1 + 3i )] ·[(1 − 3i)/(1 − 3i)]
 z = [(7 − 21i + 8i − 24i^{2})/(1 + 9)] = [(7 − 13i + 24)/10]
 (1 − 3i)(1 − 3i)(1 − 3i)
 (1 − 3i − 3i + 9i^{2})(1 − 3i)
 ( − 8 − 6i)(1 − 3i)
 − 8 + 24i − 6i + 18i^{2}
 (1 − 2i)(1 − 2i)(1 − 2i)
 (1 − 2i − 2i + 4i^{2})(1 − 2i)
 ( − 3 − 4i)(1 − 2i)
 − 3 + 6i − 4i + 8i^{2}
 (1 + 3i)(1 + 3i)(1 + 3i)
 (1 + 3i + 3i + 9i^{2})(1 + 3i)
 ( − 8 + 6i)(1 + 3i)
 − 8 − 24i + 6i + 18i^{2}
 i^{ − 7} = i^{ − 3} = i
 i^{23} = i^{3} = − i
 i^{30} = i^{2} = − 1
 i^{40} = i^{0} = 1
 i^{50} = i^{2} = − 1
 i^{25} = i^{1} = i
 ( 4 + [5/(4 − 2i)]·[(4 + 2i)/(4 + 2i)] )^{2}
 ( 4 + [(20 + 10i)/20] )^{2}
 ( 4 + 1 + [i/2] )^{2}
 ( 5 + [i/2] )^{2}
 ( [10/2] + [i/2] )^{2}
 ( [(10 + i)/2] )^{2}
 [(100 + 20i + i^{2})/2]
 ( 2 + [4/(3 + 2i)]·[(3 − 2i)/(3 − 2i)] )^{2}
 ( 2 + [(12 − 8i)/13] )^{2}
 ( [26/13] + [(12 − 8i)/13] )^{2}
 ( [(14 − 8i)/13] )^{2}
 ( 3 + [1/(2 + i)]·[(2 − i)/(2 − i)] )^{2}
 ( 3 + [(2 − i)/5] )^{2}
 ( [15/5] + [(2 − i)/5] )^{2}
 ( [(13 − i)/5] )^{2}
 [(169 − 26i + i^{2})/25]
*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 0:00
 Main Definition 0:07
 Number i
 Complex Number Form
 Powers of Imaginary Number i 1:00
 Repeating Pattern
 Operations on Complex Numbers 3:30
 Adding and Subtracting Complex Numbers
 Multiplying Complex Numbers
 FOIL Method
 Conjugation
 Dividing Complex Numbers 7:34
 Conjugate of Denominator
 Example 1: Solve For Complex Number z 11:02
 Example 2: Expand and Simplify 15:34
 Example 3: Simplify the Powers of i 17:50
 Extra Example 1: Simplify
 Extra Example 2: All Complex Numbers Satisfying Equation
Trigonometry Online Course
I. Trigonometric Functions  

Angles  39:05  
Sine and Cosine Functions  43:16  
Sine and Cosine Values of Special Angles  33:05  
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D  52:03  
Tangent and Cotangent Functions  36:04  
Secant and Cosecant Functions  27:18  
Inverse Trigonometric Functions  32:58  
Computations of Inverse Trigonometric Functions  31:08  
II. Trigonometric Identities  
Pythagorean Identity  19:11  
Identity Tan(squared)x+1=Sec(squared)x  23:16  
Addition and Subtraction Formulas  52:52  
Double Angle Formulas  29:05  
HalfAngle Formulas  43:55  
III. Applications of Trigonometry  
Trigonometry in Right Angles  25:43  
Law of Sines  56:40  
Law of Cosines  49:05  
Finding the Area of a Triangle  27:37  
Word Problems and Applications of Trigonometry  34:25  
Vectors  46:42  
IV. Complex Numbers and Polar Coordinates  
Polar Coordinates  1:07:35  
Complex Numbers  35:59  
Polar Form of Complex Numbers  40:43  
DeMoivre's Theorem  57:37 
Transcription: Complex Numbers
We are working on some examples involving arithmetic with complex numbers and we have a complicated expression to simplify here 4+2/3I all quantity squared.0000
The first thing we need to do here is perform a division and remember that the way you divide complex numbers is you multiply by the conjugate.0014
The point of that is if you have x +y(i) and you multiply it by its conjugate xy(i), then those multiplied together using the different of squares formula.0024
That is x^{2}  y(i)^{2} but i^{2} is 1 so this is really x^{2} + y^{2}.0036
That is the way of turning a complex number into a plain old real number.0046
Here we are going to write 4 + 2, now I’m going to multiply top and bottom by the conjugate of 3i.0053
At the end we have to square the whole thing, so this is (4 + 2/3i)(3+i/3+i) and this is 4 + 2.0070
The point of multiplying by the conjugate is that (3i)(3+i) using this difference of squares formula is 3^{2} + 1 ^{2} .0089
And then in the numerator 2 x (3+ i) is 6 + (2i), we still have to square it so this is 4 + (6+2i), 3^{2} + 1 ^{2} that is 9 + 1=10 squared.0104
I can simplify the fraction a little bit, we can write that as 4 + 3 + (i)/5.0124
I’m just taking out 2 out of everything there and now I’m going to put it over a common denominator.0132
That is 20 + 3 + i/(5^{2}) that is 23 + i/(5^{2}).0144
Now remember that (a + b)^{2})is a^{2} + 2ab + b^{2}.0166
23^{2} is 529 + 2ab 2(23)(i)that is 46i + i^{2} that is 1, then 5^{2} is 25 .0176
So this whole thing simplifies down to 528 + 46i/25.0196
There are several things that made this problem work, the first thing to remember is when you can basically add, subtract, multiply, and divide complex using the same rules as real numbers.0212
But when you want to divide a complex numbers, the way you to do it is to multiply the top and bottom by the conjugate of the complex number.0222
That is why when have (3i) in the denominator, we multiplied it by the conjugate, we multiply it by 3 + i.0232
That lets you exploit this formula, this x^{2} + y^{2} formula, that is why I get 3^{2} + 1 ^{2} in the denominator.0243
Then I got 10 in the denominator, I did a little bit of fraction simplification to get down to this stage 23 + i/5.0252
I expanded out the square using (a+b)^{2} = a^{2} + 2ab + b^{2}.0262
That is how I get to this line remembering that i^{2} is 1 and then I simplified it down to get the answer.0270
Finally we have to find all complex numbers satisfying z^{2}= 3 + 4(i), this is going to be a little bit complicated.0000
We are going to try to imagine that we have a complex number (x+y)^{2}=3+4(i) with (x) and (y) being real.0009
We will expand that out and we will see what happens.0037
(x+y)^{2} is x^{2} + 2xy(i) + yi^{2} but that is the same as y^{2} since (i)^{2} is 1.this is equal to 3 + 4i.0039
If I equate the real and the imaginary parts here, I will get the real part on each side is x^{2} y^{2}=3.0062
The imaginary part I will write this in red is 2(xy)(i)=4(i), that tells me that 2(xy) is equal to 4, we can simplify down a little bit I get (xy)=2.0076
This is really two equations and two unknowns, I have x^{2}  y^{2}=3 and xy=2.0098
A little bit harder than the two equations and two unknowns that you usually study in algebra class though because this is not linear.0105
To be a little more work to solve these, they are not linear equations, they are not equations to form ax + by=c, but we can still solve them.0112
I think I’m going to solve the bottom equation for y, so y=2/x and we are going to take that and plug it into the top equation and so I get x^{2}  (2/x)^{2} = 3.0121
Now I narrowed it down to one equation and I will try to solve for that on variable x^{2}  4/x^{2} = 3.0148
We got to multiply both sides here by x^{2} to clear my denominator.0160
I get( x^{4}  4) =3(x) ^{2}, I will move the 3(x) ^{2} over so I get x^{4}  3(x) ^{2} 4=0.0170
That looks a lot like a quadratic equation except that it goes up to the 4th degree.0185
But there are no odd powers of (x) in there, I just have x^{2} and x^{4}.0191
I’m going to make a little substitution, I’m going to let w=x^{2} and the point is that is we have not done yet x^{4}=w^{2}  3w  4=00197
That is really nice because that is now a quadratic equation in (w), I know how to solve that.0217
I can use a quadratic formula, I can complete the square, but I’m going to factor because I think that is the easiest.0221
(w4) x (w+1) = 0 and (w)=4 or (w)=1.0229
Now I will substitute back in now to get things back in terms of x, I get x^{2}=4 or x^{2}=1.0247
You might think “I know something that has a perfect square equal to 1 that is (i)”.0260
But remember we said x and y are real numbers so we can not have x^{2}= 1 since (x) is a real number.0265
That x^{2}=1 really does not make sense and it must be x^{2}=4.0295
I’m going to the next slide, we had our solution so far was (z)=x+y(i) with (x) and (y) being real numbers.0300
The way what we solved so far was we got down to x^{2}=4, so x=2 or x=2.0331
If you remember back on the previous side, we had solved that y=2/x, for each one of these (x) values we get a corresponding (y) value.0343
If x=2 then y=2/2 that would be 1, if x=2 y would be 2/2 so we get y would be 1.0354
This would give me (z)=2+(i) because (z) was x + y(i), this solution would give me z = 2 – (i), those are the two solutions.0368
I like to check that quickly just to make sure it works because there was a lot of work to get through there, I want to make sure I did not do any mistakes.0393
We are going to plug that in, z^{2} would be 2 + (i) ^{2} and actually notice that the second solution is just the negative of the first solution.0403
If you square a negative number, the negative just goes away.0421
We really only have to check one because if the first one is right, then the second one will also be right because it is the negative of the first one.0427
This is 2^{2} + ((2 x 2 x (i)) that is 4(i) + (i) ^{2}.0434
I’m filling it out or I’m using the old algebra formula (a+b) ^{2} is a^{2} + 2ab + b^{2}.0443
This is 4 + 4(i), (i) ^{2} is 1, this is 3 + 4(i), that does indeed check there.0453
Let me recap the strategy that we used there, we guessed z=((x+y(i)).0466
We plugged it in to (x+y)(i) ^{2} = the original equation 3 + 4(i).0472
We expanded out (x+y)(i) ^{2} using foil into something, a real number + a real number x (i).0483
That is equal to 3 + 4(i), we can equate the coefficients on each sides.0496
We said this one must be equal to 3, and this one must be equal to 4.0501
That gave us two equations in (x) and (y) and although they are not linear equations it was a little extra work to solve them.0507
We could solve them down, I believe we set it to one another and we got a 4th degree equation in (x).0514
To solve that we let w=x^{2}, and we got a quadratic equation in (w).0524
We are able to factor that into two solutions in terms of (w).0535
Each one of those converted into a solution for (x) but when we got x^{2}= something or x^{2}= something else one of those was 1.0543
That is not legitimate for a real numbers, that is only legitimate for complex number, (x) was supposed to be real.0558
We threw out that one, the other one was x^{2}=4, that gave possibilities of 2 or 2 for (x).0567
Each one of those gave me a corresponding lie, remembering our original substitution and so we got our two solutions for (z).0575
We could check each one by squaring them out and see if we really get 3 + 4(i), I just squared up the positive one because I know that if you square the negative it will give you the same answer there.0585
That is the end of our lecture on complex numbers, these are the trigonometry lectures on www.educator.com.0595
Hi this is Will Murray for educator.com and we're here today to learn about complex numbers.0000
Up until now, when you have been studying math, you've always learned that you can only take the square root of a positive number.0008
Complex numbers, the idea is, we're going to allow ourselves to take square roots of negative numbers.0015
We'll start by creating this number that we call i.0023
The rule for i is that i^{2}=1.0029
Now, we'll talk about complex numbers which are numbers of the form (x+y)i.0033
We're going to talk about adding and subtracting those, and multiplying.0042
We'll learn how to multiply and divide those.0045
Just remember that x and y are real numbers, and i always satisfies this rule, that i^{2} is 1.0048
Now we can talk about the square root of a negative number.0055
We'll start off by looking at powers of i because they do follow a pattern i^{0}, just like anything else to the 0, we define to be 1, we say 1^{0} is 1.0060
The first power is i, i^{2}=1, i^{3}=i^{2}×i, remember i^{2}=1, so i^{3}=i, i^{4} remember is i^{2}×i^{2}=1×1, i^{4}=1.0072
Since we've come back to 1, if we multiply on one more power of i, i^{5} is just i again.0102
You could see that we get this repeating pattern, 1, i, 1, i.0110
This pattern keeps on repeating, you can figure out any power of i just by remembering this pattern.0119
In fact, it keeps going in the other direction too.0126
Let me write down some of the negative powers of i, we'll be using them later.0128
If you go back in the other direction, i^{1}, well it's 1 power before you get to 1, and if you read back from the bottom, i^{1}, 1 power before you get to 1 would be i.0131
i^{2} would be 1.0153
i^{3} would be i.0156
i^{4} would be 1.0158
i^{5} would be i.0163
i^{6} would be 1.0170
i^{7} would be i.0178
i^{8} would be 1 again.0181
If you just follow the powers then they go in this pattern of fours, 1, i, 1, i, 1, i, 1, i.0186
Just remember they go by fours and they always follow that pattern 1, i, 1, i.0202
Let's learn how to do operations on complex numbers, addition and subtraction and then the other ones are more complicated.0210
The addition one's pretty easy.0217
If you want to add x+yi to another complex number a+bi, you just add the real parts together and then the imaginary parts.0220
The real parts are the x and the a, you just add those together to get x+a, then the imaginary parts, I'll do those in red, yi+bi, you add those together to get (y+b)i.0229
Subtraction is very much similar.0249
You just subtract the real parts from each other, then subtract the imaginary parts.0251
x+yi(a+bi), first you subtract the real parts, you get xa, then you subtract the imaginary parts yibi is (yb)i.0255
We'll practice this with actual complex numbers when we get to the examples.0271
You have plenty of time to practice these with actual numbers.0274
Multiplication is more complicated and it's not what you might think.0279
You might think that you just multiply the real parts, and you multiply the imaginary part.0284
It's not like that.0289
What you have to do here is FOIL it out.0291
Let me show how that works a little more slowly.0293
If you want to do (x+yi)×(a+bi), remember in the algebra lectures that you have to FOIL it out, first inner, outer last.0296
Let me write that out, first inner, outer last.0309
The first two terms are x×a, the inner terms are yai, the outer terms are xbi, then the last terms are yb, then i^{2}.0311
If you simplify that, you get x×a, the ybi^{2}, remember i^{2} is 1, that's xayb.0333
I'm going to factor an i out of the other two terms because those are my imaginary terms, (xb+ya)i.0346
For the answer, the real part is this xayb, then the imaginary part is (xb+ya)i.0357
Now you can see where this complicated formula comes from.0366
It just comes from Foiling out the four terms there.0371
There's another common operation that you need to learn about what complex numbers called conjugation.0375
You write the conjugate of a complex number with a bar over it, like this.0382
Conjugation means you just take x+yi, and you change it into xyi, or if it's xyi, you change it into x+yi.0389
Conjugation is very useful because if you multiply a complex number in its conjugate, x+yi times it's conjugate xyi, you get this difference of squares formula x^{2}(yi)^{2}.0401
But x^{2}, remember i^{2} is 1, this turns into x^{2}+y^{2}.0421
That's very useful when you're trying to produce a real number one would know imaginary part, no i part.0429
You multiply a complex number by its conjugate, and the answer would always be this real number, in fact, it'll always be a positive real number because x^{2}+y^{2} will always be positive.0436
Now we're ready to learn the most complicated operation on complex numbers which is division.0449
How do you divide x+yi by a+bi?0454
We want to give our answer in a real number plus a real number times i.0457
This is quite complicated.0464
The trick is to look at the conjugate of the denominator, a+bi conjugate, and multiply top and bottom by the conjugate.0465
That's what we'll do here, we multiply top and bottom by a+bi conjugate.0474
The conjugate of a+bi is abi.0481
Remember that the point of multiplying a complex number by its conjugate is that, when you multiply (a+bi)×(abi), the answer is just a real number a^{2}+b^{2}.0485
That's where the denominator comes from, it comes from that difference of squares pattern we learned from the previous slide, (a+bi)×(abi)=a^{2}+b^{2}.0500
In the numerator, it doesn't work so nicely,we have to multiply out (x+yi)×(abi).0510
If we FOIL it out again, we have x×a first terms, outer terms are xbi, inner terms are +yai, then the last terms are ybi^{2}.0517
If we simplify that and collect the real terms, remember i^{2} is 1, this is +yb, if we combine that with the xa, that's where this real terms comes from xa+yb.0536
If we combine this 2 imaginary terms and factor out the i, we get the (yaxb)i.0554
That's where that line comes from.0560
You can separate this out to get it into the form we're looking for.0561
Remember we wanted to form a real number plus a real number times i.0566
If you separate this out, we get (xa+yb)/(a^{2}+b^{2}), that's a real number, then ((yaxb)/(a^{2}+b^{2}))i.0572
We got it into that form we liked.0583
That seems extremely complicated.0587
The only thing you need to remember here is to divide complex numbers ...0590
I'll write this down, to divide complex numbers, the only thing you need to remember is multiply top and bottom by the conjugate of the denominator.0603
That's what we did here.0632
We multiplied top and bottom by abi which came from the denominator being a+bi.0633
You just need to remember to multiply the top and bottom by the conjugate of the denominator.0643
The point of that is that makes the denominator a^{2}+b^{2} and the numerator you kind of stuck with whatever mess you end up getting into.0648
Let's practice now with some actual complex numbers and see how that works.0658
We want to solve the following equation for the complex number z.0663
Our unknown is z.0668
Think of this as being a complex number times z, plus a complex number is equal to another complex number.0670
This is sort of like solving with these real numbers az+b=c, then we would solve for z by subtracting b from both sides az=cb, divide both sides by a, and we get z=(cb)/a.0676
We're going to do the analog of that process but with complex numbers.0699
First, we'll subtract from both sides 6+2i.0702
We'll subtract that from both sides.0713
That will give us on the left (1+2i)×z=(2+9i)(6+2i).0714
That simplifies down to 4+7i.0728
Now we'll divide both sides by 1+2i, I'm trying to solve for z.0736
I get z=(4+7i)/(1+2i).0747
Remember the way you do division for complex numbers, you multiply top and bottom by the conjugate of the denominator.0758
I'll multiply this by 1+2i conjugate.0766
This is (4+7i)/(1+2i).0776
The conjugate of 1+2i is 12i.0782
The point of that multiplication is it makes the denominator very simple.0791
It's 1^{2}(2i)^{2} but since i^{2}=1, this is just +2^{2}.0798
Remember that's because i^{2}=1.0810
The numerator is going to get more messy and there's no way to get around expanding this using FOIL.0813
I've got 4, first terms, the outer terms are +8i, the inner terms are 7i×1, the last terms are 14i^{2}, but i^{2} is 1, that's really +14.0819
If I simplify this down, my denominator is 5, 4+14, is 10, 8i+7i=15i, this simplifies down to 2+3i.0843
My z value there is 2+3i.0864
Let's recap what we had to do to make this problem work.0868
Essentially, think of each of these complex numbers as real numbers and treat them just as real numbers, and do whatever algebraic operations you need to do to solve the equation.0872
In this case, to solve the equation we had to subtract a complex number from both sides.0883
I know how to subtract complex numbers now.0890
Then we had to divide both sides by the complex number 1+2i.0892
That's trickier.0898
To divide complex numbers, you multiply by the conjugate.0899
We multiplied by the conjugate here, that's 12i.0902
The point of multiplying by the conjugate is that 1+2i and 12i multiply using the difference of squares pattern into 1+2^{2}.0905
The reason it's plus is because we have an i^{2} which makes a minus into a plus.0917
We get a nice denominator there, the numerator we just had to expand it out using FOIL, and it simplifies down to the complex number that we found.0923
For our next example, we have to expand and simplify 1+2i^{3}.0935
I'll write that as (1+2i)×(1+2i)×(1+2i).0941
I'll just multiply these first two terms together.0952
That's (1+2i)^{2}, I'm using now (a+b)^{2}=a^{2}+b^{2}+2ab.0955
The 2ab is 2×2i, that's 4i.0975
All this times 1+2i.0981
Now, (2i)^{2}, that's 4×1, that's 4.0984
This whole thing is (3+4i)×(1+2i).0991
I'll expand that out using FOIL, first terms 3, outer terms is 3×2i, that's 6i, inner term's +4i, the last term's +8i^{2}.1003
The 8i^{2} converts into 8.1024
So, 38=11, 6i+4i=2i.1029
What we had to do for that problem, we had to cube 1+2i, that's just a matter of multiplying out 1+2i times itself 3 times.1043
We just do that in stages, we multiply two of them together using FOIL, simplify it down to a simpler complex number 3+4i, then multiply on the third one using FOIL, then simplify it down to a simpler complex number.1053
The third example here, we have to simplify the following powers of i, i to the 6, 19, 33, 13.1071
The key thing here is to remember that the powers of i go by 4.1080
If we start at i ^{0}, that's 1, i^{1}=i, i^{2}=1, i^{3}=1, after that they start repeating.1084
i^{3}=i, i^{4}=1.1099
After that they start repeating in cycles of 4, and it goes back in the other direction too.1107
i^{1}=i, i^{2}=1, i^{3}=i, i^{4}=1.1115
It goes in cycles of 4 in both directions.1131
What we have to do is look at this exponent and try to reduce them down by 4's.1135
i^{6}, I'll reduce that down by 4, that's the same as i^{2}, and I check over here, i^{2}=1.1141
i^{19}, if you're counting by 4's, you got 4, 8, 12, 16 and then 3 more to get to 19, the remainder's 3.1154
That's the same as i^{3} because we went down.1168
Essentially what we did was we went down by 4 at a time, and i^{3} I remember from my pattern over here is i.1171
i^{33}, if you're counting by multiples of 4, you have 4, 8, 12, 16, 20, 24, 28, 32, 33 is one more, that's i^{1}.1183
It cycles every 4 powers of i.1208
This i^{1} is just i.1212
i^{13}, counting by powers of 4 you got 4, 8, 12, then 13 is one more than 12, this is the same as i^{1}.1216
We know that that's i.1229
The way you simplify powers of i is you just remember that they go in multiples of 4, they go in cycles of 4.1240
If you have a really big power of i, you just need to figure out what power is, what its remainder is when you divide by 4.1250
For example in the case of 19, 19=4×4+3, you throw out the 4's and just look at that remainder of 3, you get i^{3}, then you remember i^{3}=i.1257
It works for all of them the same way.1278
You just have to keep track of what the cycle is for the first four powers of i.1280
2 answers
Last reply by: Dr. William Murray
Fri Aug 16, 2013 6:24 PM
Post by Suhani Pant on August 16, 2013
In extra example #1, you said that the denominator is 3 squared plus 1 squared. However, I thought multiplying by the conjugate would result the answer to be 3 squared plus i squared which is 3 squared minus 1. So,wouldn't the denominator end up being 8?
2 answers
Last reply by: Jorge Sardinas
Wed Jun 12, 2013 10:38 AM
Post by Jorge Sardinas on January 24, 2013
in the conjugation instead of x^2+y^2 it should be x^2y^2,just saying!
3 answers
Last reply by: Dr. William Murray
Wed Apr 3, 2013 11:43 AM
Post by Norman Cervantes on December 9, 2012
it wasnt till the second time that it all started to make sense, i just wish i wouldve known about this website back in high school!
1 answer
Last reply by: Dr. William Murray
Mon Dec 10, 2012 11:24 AM
Post by varsha sharma on June 8, 2011
For example 2 , we could use (a+b)cubed identity, and expand directly.
Varsha
1 answer
Last reply by: Dr. William Murray
Mon Dec 10, 2012 11:21 AM
Post by David Rodriguez on November 8, 2010
instructor said at 5:07 (FIOL)firs inner outer last, should of said (FOIL)First outer inner last. Just saying