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## Precalculus

### Course: Precalculus > Unit 3

Lesson 5: Modulus (absolute value) and argument (angle) of complex numbers- Absolute value of complex numbers
- Complex numbers with the same modulus (absolute value)
- Modulus (absolute value) of complex numbers
- Absolute value & angle of complex numbers
- Angle of complex numbers
- Complex numbers from absolute value & angle
- Complex number absolute value & angle review

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# Absolute value & angle of complex numbers

Sal finds the modulus (which is the absolute value) and the argument (which is the angle) of √3/2+1/2*i. Created by Sal Khan.

## Want to join the conversation?

- I don't understand how e just popped into the equation. Is there another video series I should be referencing to understand where this came from? I've already watched the compound interest series that introduces e but that didn't prep me for this.(118 votes)
- This video should be in Calculus playlist, not in Precalculus. Because you're not supposed to have even heard of Euler's formula if you are learning precalculus(149 votes)

- we know real infinity(positive.negative) but is there a complex infinity?(23 votes)
- Excellent question!

Yes there is, in the stereographic mapping of the Riemann Sphere to the complex plane. The "Point at Infinity" corresponds to the "North Pole" of the Riemann Sphere. Link: http://en.wikipedia.org/wiki/Complex_plane(14 votes)

- At12:30, Sal states that imaginary numbers in exponential form should be measured in radians. Why?(11 votes)
- Degrees are a contrived unit, radians are not. From this point on, you will be usually dealing with radians not degrees. When you move on in math, you won't be having degrees at all, just radians.

The reason for considering an exponent with an imaginary unit as an angle is because of this relationship:

e^(i*x) = cos x + i sin x

(this only works if x is in radians)(20 votes)

- At8:56Sal writes +1, even though i^2=-1

Am I wrong or right?(8 votes)- He calculated the absolute value of
**z**, |**z**|, where you square the*real*parts of**z**, and then add them and take the square root.

So,

if**z = a + bi**

then the real parts are**a**and**b**

In this case**z = √(3)/2 + i**

Then**a = √(3)/2**

and**b = 1**, because the real part of**i**is**1**, just as the real part of**2i**is**2**

The absolute value of z is:

|**z**| = √(**a^2 + b^2**)

Which gives:

|**z**| = √(**3/4 + 1**)

Hope that helped!(18 votes)

- What if the complex number is not on the z=a+bi form? For example simply -2i. If I then wanted to find the argument, wouldn't I then end up with phi=atan(-2/0), which is undefined?(2 votes)
- Actually, -2i IS in the z=a+bi form. It just means a=0. Just had that one on a trig test about a month ago.(15 votes)

- In the practice questions that follow, when the angle measure must be given between -180 degrees and 180 degrees, how do you know when to add or subtract 180 degrees from the result of taking the inverse tangent to get the final answer?(5 votes)
- First, figure out which quadrant the point
**a + bi**lies in. You can do this by thinking about the signs of**a**(the real part) and**b**(the imaginary part):

- Quad I: a is + , b is +

- Quad II: a is - , b is +

- Quad III: a is - , b is -

- Quad IV: a is + , b is -

Next, work the problem and get the result.

Finally, think about which quadrant your final answer**should**be in - this will be the**same**quadrant that the original point lies in. Remember that instead of a circle that goes from 0° to 360°, we are starting at -180° (the negative x-axis... or Re-axis) and traveling**counter-clockwise**to +180°.

Now, check to see if your result is in the same quadrant as the original point**a + bi**. If not, just add or subtract 180° to get it into the correct quadrant.

You can double check to make sure that adding or subtracting 180° (or doing nothing) gives you an answer in the same quadrant as the original point.

- Quad I will be 0° to 90°

- Quad II will be 90° to 180°

- Quad III will be -180° to -90°

- Quad IV will be -90° to 0°

Hope this helps!(10 votes)

- wait so for the example you gave at the 9 minute mark where z=sqrt(3)/2 + i, when calculating r = |z|= sqrt((3/4) + (i^2)) wouldnt i^2 equal -1 or (-4/4) as oposed to it equaling 1 or (4/4). since i^2 is equal to -1 according to the imaginary number system ?(7 votes)
- Ok I think I just figured it out,correct me if Im wrong but because we're dealing with |z| = | sqrt(3)/4 +i |, we're only concerned with finding the magnitude of r = |z| in other words the distance or "hypotenuse" therefore if Re(z) or Im(z) is negative in other words if a = sqrt(3)/2 or b = i are negative we simply ignore the minus signs since we're trying to find the distance r =|z|..... another way of looking at it is if we're trying to use the Pythagoras formula r^2 = x^2 + y^2, we ignore the minus signs or in other words we ignore the quadrant that the right angled triangle is formed and calculate the magnitude of the hypotenuse whether x= negative number or y = negative number giving us a reuslt thats alway positive i.e. |r^2| =|x^2| + |y^2| ... i hope this helps anyone who had the same confusion(5 votes)

- how do i write: -1+√3 i in polar form(3 votes)
- precalculus video quoting taylor series? something is wrong folks (7:00)(4 votes)
- I'm trying to skip algebraiii/trigonometry in my school since I already know most of the material. In my textbook it uses r cis theta.

No where it says re^i(phi). What is the correlation between these two? When are they used?

Thanks!(2 votes)- cis theta is just shorthand for cos theta + i sin theta
`c`

os theta +`i s`

in theta

so, r cis theta just means r times cis theta and is therefore the same as rcos theta+risin theta or r(cos theta + i sin theta)

I'm not so fond of the cis notation because it obscures the really cool things about how sin and cos and theta can be related as complex numbers, but that is just my preference--I love to see how things tick.

Usually re^i(phi) form is taught in calculus or perhaps introduced in precalc, with more about logarithms and transcendentals. Once you get into logarithms, the concept of`re^i(phi)`

becomes a lot easier and it becomes second nature to convert back and forth. It is the exponential form of the same relationship. It is used to find roots of functions, and is a shorthand way of expressing a complex number, for example in calculus. It offers a format that is a lot more promising for getting to a solution because you can use a lot more tools to manipulate it. Natural logs are to the base e for example.

One other thing that is different in your question is the phi and the theta. I am guessing you are not asking about that--phi is often used for the arguments in physics and in Argand diagrams, but theta is also used. If two angles are given, theta is the first by convention and phi is second fiddle.

If you want to skip algebra and trigonometry you may run into several examples of this-- where your textbook has simplified things that you will have to get used to in their more conventional form. It would be good to watch as many of the advanced math videos on Khan Academy as you can to be able to encounter some of these alternate forms.(5 votes)

## Video transcript

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.