Basic complex analysis | Imaginary and complex numbers | Precalculus | Khan Academy
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Basic complex analysis | Imaginary and complex numbers | Precalculus | Khan Academy


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.

About James Carlton

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89 thoughts on “Basic complex analysis | Imaginary and complex numbers | Precalculus | Khan Academy

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

  2. @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.

  3. @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.

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

  5. @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.

  6. @3809ault Perhaps you could do some videos like that, and put it together free in Internet without repeat yourself!.

  7. 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.

  8. @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.

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

  10. 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.

  11. @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!

  12. @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!

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

  14. 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 )

  15. 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.

  16. 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.

  17. 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.

  18. 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.

  19. 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.

  20. 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

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

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

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

  24. 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

  25. 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

  26. 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??

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