What I want to do

in this video is make sure we’re comfortable with

ways to represent and visualize complex numbers. So you’re probably

familiar with the idea. A complex number,

let’s call it z– and z is the variable we do tend to

use for complex numbers– let’s say that z is

equal to a plus bi. We call it complex

because it has a real part and it has an imaginary part. And just so you’re

used to the notation, sometimes you’ll see

someone write the real part, give me the real part of z. This is a function, that

you input a complex number, and it will output the real

part, and in this case, the real part is equal to a. And you could have

another function called the imaginary part of z. You input some

complex number it’ll output the imaginary

part, or it’ll say how much are scaling

up i, and in this case, it would be b. This is a real number,

but this tells us how much the i is scaled

up in the complex number z right over there. Now, one way to visualize

complex numbers, and this is actually a very

helpful way of visualizing it when we start thinking about

the roots of numbers, especially the complex roots, is using

something called an Argand diagram. So this is this. And so it looks a lot

like the coordinate axes and it is a coordinate axes. But instead of having

an x and y-axis it has a real and

an imaginary axis. So in the example of

z being a plus bi, we would plot it really as

a position vector, where you have the real part

on the horizontal axis. So let’s say this is a and

then the imaginary part along the vertical axis,

or the imaginary axis. So let’s say that this is b. And so we would represent,

in an Argand diagram, the vector z as a position

vector that starts at 0 and that has a tip at

the coordinate a comma b. So this right here is

our complex number. This right here is

a representation in our Argand diagram

of the complex number a plus bi, or of z. Now when you draw it this way,

when you draw it as a position vector, and if you’re familiar

with polar coordinates, you’re probably

thinking, hey, I don’t have to represent this

complex number just as coordinates, just

as an a plus bi. Maybe I could represent

this as some angle here, let’s call that angle phi,

and some the distance here, let’s call that r, which

is kind of the magnitude of this vector. And you could. If you gave some angle

and some distance, that would also specify this

point in the complex plane. And this is actually called the

argument of the complex number and this right here is called

the magnitude, or sometimes the modulus, or the absolute

value of the complex number. So let’s think about

it a little bit. Let’s think about

how we would actually calculate these values. So r, which is the

modulus, or the magnitude. It’s denoted by the magnitude

or the absolute value of z1. What’s this going to be. Well, we have a

right triangle here. This side is b, length b. The base right

here has length a. So to calculate r, we can just

use the Pythagorean Theorem. r squared is going to be equal

to a squared plus b squared. Or r is going to be equal to the

square root of a squared plus b squared. If we want to figure

out the argument, this is going to

be equal to what? So let’s think about

this a little bit. We have b and a. So what trig function deals

with the opposite side of an angle and

the adjacent side? So let me write all of, let me

write the famous sohcahtoa up here. “Soh-cah-toa.” Tangent deals with

opposite over adjacent. So the tangent of

this angle, which we called the argument of the

complex number, the tangent of the argument is going to

be equal to the opposite side over the adjacent side. It is equal to b/a. And so if we wanted to

solve for this argument, we would say that

the argument is equal to the arctan, or the

inverse tangent, of b/a. Now, if we wanted

to represent, let’s say that we were given the

modulus and the argument. Let’s say we were given that. How do we go the other way? Right now if we have the a’s and

the b’s, the real complex part, I just showed you how

to get the magnitude and how to get the

angle, or the argument. But if you’re given this. How do you go the other way? Well here, if you’re

trying to figure out a, given r and theta–

so you’re trying to figure out an adjacent side

given angle and the hypotenuse. So adjacent over hypotenuse

is equal to cosine. So you would have

cosine of the argument is equal to the adjacent

over the hypotenuse. It is equal to a/r. Multiply both sides by r,

you get r cosine of phi is equal to a. Do something very similar for b. If we use sine, opposite

over hypotenuse. Sine of the argument

is equal to b/r. It is equal to b

over the magnitude. Multiply both sides by

r, you get r sine of phi is equal to b. So how would we write

this complex number. So this complex

number z is going to be equal to it’s

real part, which is r cosine of phi plus

the imaginary part times i. Plus r– let me do that

same green– plus r sine of phi times i. Now this might pop out

at you as something that’s pretty interesting, if

you ever seen Euler’s formula. Let’s factor out

this r over here. So this is going to be equal

to– factor out an r– r times cosine of phi plus– I’ll

put the i out front– i sine of phi. Now what is this? And if you’ve seen

the video, I do it in the Taylor series,

a series of videos in the calculus playlist. And it’s really one of the

most profound results and all of mathematics, it

still gives me chills. This is Euler’s formula. Or this, by Euler’s

formula, is the same thing. And we show it by looking

at the Taylor series representations of e to the x. And the Taylor series

representations of cosine of x and sine of x. But this is, if we’re dealing

with radians, e to the i phi. So z is going to be equal

to r times e to the i phi. So there’s two ways to

write a complex number. You could write it

like this, where you have the real and

imaginary part, that’s maybe what we’re used to. Or we can write it in

exponential form, where you have the modulus,

or the magnitude, being multiplied by a

complex exponential. And we’re going to see that

this going to be super useful, especially when we’re

trying to find roots. Now just to make this

tangible, let’s actually do this with an actual example. So let’s say that I

had, I don’t know, let’s say that I had to z1

is equal to square root of, let’s say it’s square

root of 3/2 plus i. And so we want to figure

out its magnitude, and we want to figure

out its argument. So let’s do that. So the magnitude of

z1 is going to be equal to the square

root of this squared. So this is going to be equal

to 3/4 plus 1 squared– or I should say plus 4/4. So this is going to be equal

to square root of 7/4, which is equal to the

square root of 7/2. And now let’s figure

out its argument. So if I were to draw this

on an Argand diagram, it would look like this. It’s going to be in

the first quadrant, so that’s all I

have to worry about. So let me draw it. Let me draw it like this. And so we have a situation. So it’s going to be

square root of 3, actually, let me change

this up a little bit, just so the numbers get

a little bit cleaner. Sorry about this. Let me make it a little

bit, slightly cleaner. So just so that we have a

slightly cleaner result, because we want to make our

first example a simple one. So let’s make this square

root of 3/2 plus 1/2i.. So let’s figure

out the magnitude, the magnitude here is z1

is equal to the square root of, square root

of 3/2 squared, is equal to 3/4 plus 1/2

squared is equal to 1/4,. This makes things a lot nicer. This is equal to the square

root of 1, which is 1. And now let’s think

about it, let’s draw it on an Argand diagram to

visualize the argument. So this is my imaginary axis. This is my real axis. And so this complex number

is square root of 3/2. The square root

of 3 is like 1.7. So if we have like 1, it’ll

be like right over here, someplace right over here. This is square root

of 3/2, the real part. The imaginary part is 1/2. So if this is 1, this is

1/2, the imaginary part is right over here, 1/2. And we actually also know

its length, its length, or its magnitude is 1. So how do we figure

out phi over here? We know that this side over

here is square root of 3 over– oh let me be careful–

we know that side over there is 1/2. That’s the imaginary part. And we know the base is

the square root of 3/2. So a bunch of ways

we can do this. One, you could just

do the tangent, because that involves the

opposite over the adjacent. You could say that

the tangent of phi is equal to the

opposite, is equal to 1/2 over the square root of 3/2. And then you can take the

inverse tan of both sides. So this would be the

same thing as phi being equal to the

inverse tangent, or the arctangent

of– if you multiply the numerator and

the denominator by 2, this is 1 over the

square root of 3. You could do it like that. You can also say that phi is

equal to the inverse sine of, so the sine of

phi is going to be equal to the opposite

over the hypotenuse. So sine of phi is

equal to 1/2 over 1, or phi is equal to

the arcsine of 1/2. And you could put that

into your calculator. Or you could recognize this

is a 30-60-90 triangle. This base right here, square

root of 3/2, this is 1/2, this is 1. So this angle right here

is going to be 30 degrees. And that’s just from

pattern matching from a 30-60-90 triangle. You could look at these and

also get something similar. Now I want to put

this in radian form, because whenever I use

the exponential form you want it to be in radians. So phi is equal to

30 degrees, which is the same thing as pi over 6. So if I wanted to represent

z1 in exponential form, it would be the exact same

thing as r, or its magnitude, which is 1– I’ll put the 1 out

there even though you really don’t have to– 1 times

e to the pi over 6i. And we’re done.

I dont know how to put this, but the timing of this video is *spot* on, just before my terminal exams. Thanks!

Good. Keep on going.

just a correction, Euler is pronounced like "Oiler"

This is so complex!

Another correction: the imaginary part of a complex number is just b, not bi.

@alawrence89 Nope, sorry. You are wrong. The "imaginary part of a complex number" is the real coefficient multiplying the imaginary unit i.

@alawrence89 I understand your reasons. In fact, a long time ago I put to myself the same question. Well, it looks like the expression "imaginary part" WAS used to denote bi, so being consistent with the word "imaginary"; later on, the i was dropped – maybe with the axiomatization of complex numbers as the set R² of ordered pairs (a, b) of real numbers? – thus leaving with just the coefficient of the "imaginary number" – that's correct, instead – bi.

@viewer280 I see that I made a mistake by using my other account to comment on this… but I assure you I'm the same person you sent the other comment to.

@experthe i heard it pronounced either way, i like the Youler pronounciation my self

Did Sal forget the negative signs when squaring i around 9:00 and 10:10? If he didn't then i'm quite confused.

if we didn't have 'e' the universe would implode.

@goldstein23 i is not squared but the coefficient of i is squared.in the first example which he deleted,the coefficient is 1 and in the next example the coefficient is 1/2.hope that helps.

"Yooler" is like fingernails on a blackboard for a mathematician.

thanks so much for this video~!!!

nice visual presentation and made a lot of things clearer 😀

@Menegoth You're just imagining it

Basic Complex Analysis…? Will you do the advanced ones?

@3809ault Perhaps you could do some videos like that, and put it together free in Internet without repeat yourself!.

This lecture should be labeled Complex Analysis for USAians. See Eurpopeans already know everything about all things. Only USAians (or Americans as we are sometimes misidentified) are ignorant about complex analysis. The only thing we could ever analyze are Big Macs and Cheeseburgers.

@harrytuttle777 Why do Europeans visualize we Americans as fat-asses?

@CallOFDutyMVP666

Because it is true. See a European is a transcendent being who is never wrong. If Europeans say we are fat, we are fat. The other day I thought I saw a fat German. Being a retarded USAian my eyeballs just could not make the kg – lb conversion. Whereas everyone in Europe speaks 6 different languages, and spends their free time looking for the Higgs boson at CERN, we USAians spend all our free time eating Big Mac's and beating up minorities. Just ask a European.

What i don't get is why someone thought of to initially raise the entire thing by e?

My teacher told me to watch this guy. lol

i thought i^2 was -1? or are you taking the absolute value of it cause of the distance thing

muahhhhhhh Te AMO

rrrrrready for my test,

this is so easy ,we want real analysis complex

ah! now I understand why they taught us that graph in algebra 2 trig, even though im in calculus 2 now. God its important.

so 1=e^itheta where ln1=i(pie/6) and 0=i(pie/6) in which case i is the square root of negative 1, 0 or hiding beyond logic's sense? yea I can imagine that.

so wait is r=z?

At 10:21 shouldn't that be – 1/4?

sorry i have a question

how cos something + i sin something related to exponential form?

@Vlaxirious It's called Euler's Identity, and it's proven using Taylor Polynomials of Sin, Cos, and e^x. If you watch the Calculus playlist, he proves it to you, but it's still quite mysterious in its own respects. It's just three functions that don't seem to be related, but become so deeply entangled when taken into complex analysis. I hope this answers your question!

@Ghaiyst r is the absolute value (distance from the origin) of z. r comes from polar coordinates for functions. It's found by taking the square root of (a^2+b^2) when z is in the form a+bi. hope it helps!

will you ever do videos on Quaternions and Octonions?

What software does he use to prepare the videos?

Isnt the square root of 1, +1 or -1? Either way though, thanks

Yes, but remember that he's trying to figure out the length of the position vector and lengths are always positive 🙂

can't have a negative modulus.

yeah..i have the same question!!

No because if you think about the Pythagorean Theorem:

c^2 = a^2 + b^2

Well, we are looking for the magnitude of the complex number, or in other words it's radius.

r^2 = a^2 + b^2

Therefore

r = Sqrt( a^2 + b^2 )

or

|z| = Sqrt( a^2 + b^2 )

|z| = Sqrt( 3/4 + 1/4 )

Sqrt( 1 ) = + or – 1..

The reason he chooses one in this scenario though is because he is looking for the length of the modulus. As Klatjofsky put it, lengths are always positive.

Because you're the one with the misunderstanding, not him.

No, that's YOU…

(Do you just know what a "FUNCTION" is ??)

I don't understand the point of complex numbers. I wish some folks could describe some applications.

I'm just learning this but i'm sure it has it's uses.Like a kid counting stones wouldn't have any use for fractions,or someone measuring the length of something won't have any use for negative numbers.At our level we probably won't know of any application for imaginary numbers but later we will.

i = rotation by 90 deg.

i x i = rotation by 180 deg.

i x i x i = rotation by 270 deg.

Complex numbers are used to indicate rotation.

Hope this helps.

Binnoy

Visualizing Maths

A book beyond formulas

Dedicated to you and all such similar minds searching for answers.

Just as there is east, north , west and south.

In maths ,there is 1, i,-1 and -i.

I have some videos over it on my channel.

Binnoy

Visualizing Maths

A book beyond formulas,

dedicated to you and all such similar minds searching for answers.

Complex numbers indicate rotation.

In maths, the number '1' indicates 'as it is'

The number 'i' indicates 'rotated by 90 deg'

The number '-1' indicates 'rotated by 180 deg'.

The number '-i' indicates 'rotated by 270 deg'.

Binnoy

Visualizing maths

A book beyond formulas

Dedicated to you and such similar minds searching for answers.

Yes, you are right, complex numbers indicated 'rotation". The comments above might help you.

Binnoy

Visualizing maths

A book beyond formulas.

Dedicated to you and such similar minds searching for answers

I appreciate your response.

I will be sure to check your channel out.

Really good video, this was my first introduction to the complex analysis and it makes me want to learn a lot more about it =)

lol basic complex analysis. … like saying this is simple hard math

remember this is all counterclockwise

thanks so much, im taking a n extension course at uni where they assume u know this (idon;t XD) 🙁 very clear much appreciated.

Thanks Sal, one would think this would have been covered in math classes at some point before partial differential equations.

Argand Diagram Argand Diagram Argand Diagram

it's very useful. thank you.^^

Yes, it is due to some reason counterclockwise.

Binnoy

Visualizing Maths

Amazon

Thanks!

My brain hurts.

Thank you so much! I don't know how many times you've heard this but you've seriously helped us. 😀

0:00

what an amazing video. really helped me a lot

thank you and as a muslim May Allah guide you to his way. 😛

excellent video! just try to stop repeating yourself repeating yourself so much

Please stop mispronouncing Euler so I don't end up throwing a rock through my monitor.

Excellent explanation of the complex number. Thank you, indeed.

Just to be in the safe side, this is done in Argand diagram right?

wait wait wait… so has no one picked up that i = root -1?? so that whole last calculation was thrown off? because that really bloody messed with me

Thank you !

this can be applied in adding 2d vectors 🙂

You saved me man. I can't thank you enough.

basic complex

why are u in versing the tan in 4:51

Want an algebraic vector to play with?

[(x-1)/(x+3), (x+2-i*x)/(x+3), (x+2+i*x)/(x+3)]

The distance is 1.

Phi is not necessarily the arctan of b/a if the tan of phi equals b/a…

Can you please teach me how to write like that with mouse.

Assalamu Alaikum wa Rahmatullah. Sir Khan I didn't get the part how u converted under root 3/2 to 3/4 . plz guide me to a video of yours so I can revise this

how did i power 2 become a positive number

instead of a negative number

oh god, Sal please tell me you're not pronouncing the great mathematician "yooler"…

The views to like Ratio….LOL

complex thing simply done

until now I thought a, b is in R (real number line)

in the formula z=a+bi, can b=0 and if so, don't we gain a real number?

pronounced "oiler" oh great khan

You are great SIR. Thank you for your online lesson.

at 10:07, why did you square the denominator from 2 to 4?

What is e here ? Is it equal to 2.71 ? Or is it another thing ? Who can tell it me

I can't believe Salman Khan pronounced Euler as "Yuu ler"

Ok so can anyone tell me what exactly is an argument of a complex number… I know how to find the argument and these stuffs

BUT WHAT EXACTLY IS AN ARGUMENT?? Why is range always

-pi to pi??

thank you so much..