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## Precalculus

### Course: Precalculus>Unit 3

Lesson 5: Modulus (absolute value) and argument (angle) of complex numbers

# 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
(151 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)
• At , 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)
• At Sal writes +1, even though i^2=-1

Am I wrong or right?
(9 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!
(19 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 ?
(8 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
(6 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!
(11 votes)
• how do i write: -1+√3 i in polar form
(3 votes)
• precalculus video quoting taylor series? something is wrong folks ()
(5 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.