Intro to complex analysis
Basic Complex Analysis Argand Diagram, magnitude, modulus, argument, exponential form
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- 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
- complex number -- let's call it z
- and z is the variable we do tend to use for complex number.
- Let's say that z is equal to a+bi.
- We call it complex because it has a real part
- it has a real part
- and it has an imaginary 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.
- In this case the real part is equal to a.
- And you can have another function
- called the imaginary part.
- The imaginary part of z.
- You input some complex number
- and it'll output the imaginary part
- The imaginary part -- it'll say how much you're scaling up i.
- And in this case it would be b.
- 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 to visualize them
- when we start to think about roots of numbers
- especially the complex roots
- Is using something called an Argand diagram.
- Argand diagram.
- So this is this
- And so it looks a lot like coordinate axes
- and it is a coordinate axes
- but instead of having an x and y axis
- it has a real and imaginary axis
- so the example of z being a+bi
- we would plot it really as a position vector
- where we have the real part on the horizontal axis
- so let's say that 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 zero and has a tip
- at the coordinate a, b.
- So this right here -- this right here
- is our complex number.
- This right here is a represation in an argand diagram
- of the complex number a+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 kind of coordinates
- just as an a+bi, maybe I can represent this as some angle here
- as some angle here - let's call that angle phi.
- And some 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 a 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 little bit
- Let's think about how we would actually calculate these values
- So r, which is the modulus or the magnitude
- its denoted by the magnitude or the absolute value of z1.
- What's this going to be?
- Well we have right triangle here.
- We have a right triangle here.
- This side is b, length b.
- The base right here has length a.
- So to calculate r we could just use the pythagorean theorem
- r squared is going to be 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.
- Let's say we want to figure out the argument.
- This is going to be equal to what?
- 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 the famous sohcahtoa appear.
- Sohcahtoa - 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 over a.
- And so if we wanted a solve for this argument
- we would say that the argument is equal to the arctan
- or the inverse tangent of b over a.
- Now, if we wanted to represent -- let's say we had the complex number
- Let's say that we were given the argument and the
- let's say that we were given --
- 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 of the real and 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 to 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 adjacent side
- given an angle and the hypotenuse.
- So adjacent over hypotenuse is equal to cosine.
- So you would have cosine of the argument is equal to r --
- is equal to the adjacent over the hypotenuse.
- Is equal to a over 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 over r.
- 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.
- The complex number z is going to be equal to it's real part
- which is r*cosine of phi
- r*cosine of phi plus the imaginary part times i
- plus r -- we do that same green -- plus r*sine of phi times i.
- times i -- Now this might pop out at you as something that's pretty interesting
- if you've ever seen Euler's formula.
- Let's factor out this r over here.
- So this is going to be equal to factor of an r.
- r times of cosine of phi
- cosine of phi plus -- i put the i up front --
- i sine of phi.
- Now what is this?
- And if you've seen the videos i did in the Taylor series
- Series of videos in the calculus playlist --
- And it's really one of the most profound results in all of mathematics
- it still gives me chills -- this is Euler's formula.
- This is -- or this by Euler's formula is the same thing.
- This 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
- e to the i, phi.
- So z is going to be equal to r
- is going to be equal to r times e to the i,phi
- e to the i, phi.
- So there are two ways to write a complex number
- You could write it like this, where you have the real and the imaginary part
- That's my "do 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.
- We're gonna see that this can be super useful especially
- when we're trying do 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 z1 is equal to
- square root of three over two plus i.
- And so we want to figure out.
- We want to figure out it's magnitude and
- we want to figure out it's argument.
- So let's do that. So that magnitude
- the magnitude of z1 is going to be equal to the square root of this square
- so this is going to be equal to
- this is going to be equal to three over four -- three fourths
- plus one squared -- we're gonna actually say
- plus four fourths, so this is going to be equal to
- square root of seven over four which is equal to
- the square root of seven over two.
- And now let's figure out it's argument.
- So let's draw -- if I were to draw this on an Argand diagram
- 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 three
- actually let me change this up a little bit
- just so the numbers get a little bit cleaner
- sorry about this
- let me ma make it a little bit, slightly cleaner.
- So just so that we have a slightly cleaner result.
- We want to make our first example a simple one.
- So let's make square root of three over two plus one half
- plus one half i.
- So let's figure out the magnitude
- the magnitude here -- z1 is equal to the square root of
- square three over two squared is equal to three over four
- plus one half square is equal to one fourth.
- This makes things a lot nicer.
- This is equal to the square root of one which is one.
- And now let's think about it -- let's draw it on an Argand diagram
- do visualize the argument.
- So this my imaginary axis.
- That is my imaginary axis.
- This is my real axis.
- My real axis.
- And so this complex number is square of three over two
- square of three over two is 1.7 so if we have like one
- it would be like -- so this is one -- it'll be right over here.
- Some place right over here.
- This is square root of three over two.
- The real part.
- The imaginary part is one half.
- So if this is one, this is one half, the imaginary part is right over here -- one half
- And we actually also know it's length -- or magnitude is one.
- So how to we figure out
- how do we figure out phi over here
- We know this side over here is square root of three over..
- Oh let me be careful -- we know that side over there is one half.
- That's the imaginary part and we know the base is square root of three over two.
- So a bunch of ways we can do this
- One. You could just look at -- do the tangent
- because that involves the opposite over the adjacent.
- You could say that tangent of phi is equal to the opposite.
- is equal to one half over the square root of three over two.
- And then you can take the inverse tan of both sides.
- So this is to be the same thing as
- phi being equal to the inverse tangent
- or the arc tangent of
- if you multiply the numeral denominator by 2
- this is one over the square root of three.
- You could do it like that.
- You could 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 one half over one-
- Or phi is equal to arc sine of one half -- and you could put that in your calculator.
- Or you could recognize, this is a 30-60-90 triangle.
- This base right here is square root of three over two.
- This is one half -- This is one.
- 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 radiant form,
- because whenever I use the exponential form you want it to be in radiant.
- So phi is equal to 30 degrees.
- phi is equal to 30 degrees which is the same thing as phi over six.
- So if I wanted to represent z1 in exponential form
- it would be exact same thing as r or it's magnitude which is 1.
- I put the one out there but you really don't have to
- one times e to the pi over six i
- e to the pi over six i.
- And we're done!
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