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Lecture Comments (14)

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^2-y^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.

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

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 i2 = − 1.
  • The powers of i satisfy the following pattern:
    − 1
    − 1
  • Operations:
    (x + yi) + (a+bi)
    (x +a) + (y + b)i
    (x + yi) − (a+bi)
    (xa) + (yb)i
    (x + yi)(a+bi)
    (xayb) + (xb + ya)i

    x + yi

    x + yi
    x2 + y2
  • Division:





    (xa +yb) + (yaxb)i

    a2 +b2
    (xa +yb

    a2 +b2
    ) + (yaxb

    a2 +b2

Example 1:

Solve the following equation for the complex number z:
(1+2i)z + (6+2i) = 2 + 9i

Example 2:

Expand and simplify (1+2i)3.

Example 3:

Simplify the following powers of i: i− 6, i19, i33, i− 13

Example 4:

Simplify the following expression:
(4 +2


Example 5:

Find all complex numbers z satisfying z2 = 3 + 4i.
Give me an extra page here.

Complex Numbers

Solve the following equation for the complex number z: (1 − 2i)z + (3 − 2i) = 2 − 8i
  • 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 − 12i2)/(1 + 4)] = [( − 1 − 8i + 12)/5]
z = [(11 − 8i)/5] or z = [11/5] − [8i/5]
Solve the following equation for the complex number z: (1 + 3i)z − (4 + 3i) = 3 + 5i
  • 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 − 24i2)/(1 + 9)] = [(7 − 13i + 24)/10]
z = [(31 − 13i)/10] or z = [31/10] − [13i/10]
Expand and simplify (1 − 3i)3
  • (1 − 3i)(1 − 3i)(1 − 3i)
  • (1 − 3i − 3i + 9i2)(1 − 3i)
  • ( − 8 − 6i)(1 − 3i)
  • − 8 + 24i − 6i + 18i2
− 26 + 18i
Expand and simplify (1 − 2i)3
  • (1 − 2i)(1 − 2i)(1 − 2i)
  • (1 − 2i − 2i + 4i2)(1 − 2i)
  • ( − 3 − 4i)(1 − 2i)
  • − 3 + 6i − 4i + 8i2
− 11 + 2i
Expand and simplify (1 + 3i)3
  • (1 + 3i)(1 + 3i)(1 + 3i)
  • (1 + 3i + 3i + 9i2)(1 + 3i)
  • ( − 8 + 6i)(1 + 3i)
  • − 8 − 24i + 6i + 18i2
− 26 − 18i
Simplify the following powers of i as − i, i, 1, or − 1: i − 7,i23,i30,i − 12
  • i − 7 = i − 3 = i
  • i23 = i3 = − i
  • i30 = i2 = − 1
i − 12 = i0 = 1
Simplify the following powers of i as − i, i, 1, or − 1: i40,i50,i25,i67
  • i40 = i0 = 1
  • i50 = i2 = − 1
  • i25 = i1 = i
i67 = i3 = − i
Simplify the following expression ( 4 + [5/(4 − 2i)] )2
  • ( 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 + i2)/2]
[(99 + 20i)/2]
Simplify the following expression ( 2 + [4/(3 + 2i)] )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
[(132 − 224i)/169]
Simplify the following expression ( 3 + [1/(2 + i)] )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 + i2)/25]
[(169 − 26i)/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.


Complex Numbers

Lecture Slides are screen-captured 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

Transcription: Complex Numbers

We are working on some examples involving arithmetic with complex numbers and we have a complicated expression to simplify here 4+2/3-I 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 x-y(i), then those multiplied together using the different of squares formula.0024

That is x2 - y(i)2 but i2 is -1 so this is really x2 + y2.0036

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 3-i.0053

At the end we have to square the whole thing, so this is (4 + 2/3-i)(3+i/3+i) and this is 4 + 2.0070

The point of multiplying by the conjugate is that (3-i)(3+i) using this difference of squares formula is 32 + 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), 32 + 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/(52) that is 23 + i/(52).0144

Now remember that (a + b)2)is a2 + 2ab + b2.0166

232 is 529 + 2ab 2(23)(i)that is 46i + i2 that is -1, then 52 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 (3-i) in the denominator, we multiplied it by the conjugate, we multiply it by 3 + i.0232

That lets you exploit this formula, this x2 + y2 formula, that is why I get 32 + 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 = a2 + 2ab + b2.0262

That is how I get to this line remembering that i2 is -1 and then I simplified it down to get the answer.0270

Finally we have to find all complex numbers satisfying z2= 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 x2 + 2xy(i) + yi2 but that is the same as -y2 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 x2 -y2=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 x2 - y2=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 x2 - (2/x)2 = 3.0121

Now I narrowed it down to one equation and I will try to solve for that on variable x2 - 4/x2 = 3.0148

We got to multiply both sides here by x2 to clear my denominator.0160

I get( x4 - 4) =3(x) 2, I will move the 3(x) 2 over so I get x4 - 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 x2 and x4.0191

I’m going to make a little substitution, I’m going to let w=x2 and the point is that is we have not done yet x4=w2 - 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

(w-4) 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 x2=4 or x2=-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 x2= -1 since (x) is a real number.0265

That x2=-1 really does not make sense and it must be x2=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 x2=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, z2 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 22 + ((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 a2 + 2ab + b2.0443

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=x2, 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 x2= something or x2= 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 x2=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

Hi this is Will Murray for 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 i2=-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 i2 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 i0, just like anything else to the 0, we define to be 1, we say 10 is 1.0060

The first power is i, i2=-1, i3=i2×i, remember i2=-1, so i3=-i, i4 remember is i2×i2=-1×-1, i4=1.0072

Since we've come back to 1, if we multiply on one more power of i, i5 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 x-a, then you subtract the imaginary parts yi-bi is (y-b)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 i2.0311

If you simplify that, you get x×a, the ybi2, remember i2 is -1, that's xa-yb.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 xa-yb, 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 x-yi, or if it's x-yi, 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 x-yi, you get this difference of squares formula x2-(yi)2.0401

But x2, remember i2 is -1, this turns into x2+y2.0421

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 x2+y2 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 a-bi.0481

Remember that the point of multiplying a complex number by its conjugate is that, when you multiply (a+bi)×(a-bi), the answer is just a real number a2+b2.0485

That's where the denominator comes from, it comes from that difference of squares pattern we learned from the previous slide, (a+bi)×(a-bi)=a2+b2.0500

In the numerator, it doesn't work so nicely,we have to multiply out (x+yi)×(a-bi).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 -ybi2.0517

If we simplify that and collect the real terms, remember i2 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 (ya-xb)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)/(a2+b2), that's a real number, then ((ya-xb)/(a2+b2))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 a-bi 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 a2+b2 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=c-b, divide both sides by a, and we get z=(c-b)/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 1-2i.0782

The point of that multiplication is it makes the denominator very simple.0791

It's 12-(2i)2 but since i2=-1, this is just +22.0798

Remember that's because i2=-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 -14i2, but i2 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 1-2i.0902

The point of multiplying by the conjugate is that 1+2i and 1-2i multiply using the difference of squares pattern into 1+22.0905

The reason it's plus is because we have an i2 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+2i3.0935

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=a2+b2+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 +8i2.1003

The 8i2 converts into -8.1024

So, -3-8=-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, i1=i, i2=-1, i3=1, after that they start repeating.1084

i3=-i, i4=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

i19, 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 i3 because we went down.1168

Essentially what we did was we went down by 4 at a time, and i3 I remember from my pattern over here is -i.1171

i33, if you're counting by multiples of 4, you have 4, 8, 12, 16, 20, 24, 28, 32, 33 is one more, that's i1.1183

It cycles every 4 powers of i.1208

This i1 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 i3, then you remember i3=-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