Okay, welcome to UC Berkeley,
and we are at Evans Hall-- that's the math building, because right now,
I'm going to do some math questions with you guys.
And I'm just going to find and show you guys a classroom,
and I'm going to use this classroom right here.
Let's see how it is inside,
and as you guys can see, the typical math classroom at UC Berkeley-- a lot of things have changed,
but the blackboard stays the same.
Surprisingly, there are two televisions.
Right. Cool, huh?
And I'll be using the middle board right here, of course.
And as you can see, the eraser here at UC Berkeley, is long.
Okay, we'll do something challenging this video,
and the equation we're going to solve is sin(𝑧)=2.
Is that even possible?
Well, if we're just talking about the real world, no, we have no solutions for this equation, right?
However, if we are talking about the imaginary world,
a lot of things can happen.
In fact, we do have complex solutions for this equation here,
And that's why you use 𝑧 instead of, like, 𝑥 or 𝜃, all right?
So, how are we going to deal with this?
I will have to first give you a new definition for sine,
and let's take a look of the Euler's formula first:
𝑒^(𝑖𝑧) = cos(𝑧) + 𝑖sin(𝑧),
and you see right here, the input 𝑧, right?
and this is the famous Euler's Formula.
If you want to see the proof of this you can check the video in the description.
And now, this is what I'm going to do:
I'm going to look at 𝑒^(-𝑖𝑧), all right?
So, in this case, my new input is technically -𝑧, right? -𝑧.
𝑖 is still the 𝑖. In this case,
well, we will just have cos(-𝑧) inside, that's a new input,
and then plus, 𝑖 is still 𝑖, sine is still sine, but the input is -𝑧, all right?
So, these are the two equations that I'm going to look at.
I'm going to keep the first one how it is and let me write down: cos(𝑧) + 𝑖sin(𝑧) = 𝑒^(𝑖𝑧).
And then for the second one, let me put this part first, but you know cosine is an even function,
so the negative doesn't matter, so I can just write this down as cos(𝑧) as well.
And in this case, sine is an odd function; I can put a negative outside, right?
So we can talk about this as -𝑖sin(𝑧) because we took the negative outside already.
And this right here is equal to 𝑒^(-𝑖𝑧) like this.
And now what I'm going to do is multiply this equation by a negative,
so let's put on this negative and make this positive and make this negative, all right?
So that I can just combine them together because my goal is to solve for sin(𝑧).
Cosine, cosine, they will cancel right here,
and then this right here is 2𝑖sin(𝑧) = 𝑒^(𝑖𝑧) - 𝑒^(-𝑖𝑧), right?
And then we can just divide both sides by 2𝑖 so we just have sin(𝑧) = (𝑒^(𝑖𝑧) - 𝑒^(-𝑖𝑧))/(2𝑖), all right?
So this is the new definition of sine in the complex world that we are going to use in this video.
so now let's get back to this right here.
Once again, sin(𝑧) = (𝑒^(𝑖𝑧) - 𝑒^(-𝑖𝑧))/(2𝑖),
and then we can just say this is equal to 2 on the right side,
and now we'll just be doing some typical algebra.
We can multiply two I on both sides, so this is 𝑒^(𝑖𝑧) - 𝑒^(-𝑖𝑧) = 4𝑖.
And let me tell you this is very similar to sinh (hyperbolic sine) ok? sinh(𝑥), but yeah.
sinh⁻¹(𝑥), all right? Whatever.
Anyway,
I'm going to solve for 𝑧, right? Because 𝑧 is an input, if I can get 𝑧 as a solution.
So now that we have +𝑖𝑧 here; this is -𝑖𝑧, right?
so if you would like, we could just multiply everything by 𝑒^(𝑖𝑧).
So you'll see the following:
This, distribute, and multiply, right? So the first one is (𝑒^(𝑖𝑧))²,
because we have this times that, so it's something squared, right?
And then this times that, well, the power becomes 0; we'll just have -1,
and this is equal to 4𝑖 times that-- I should do this.
4𝑖*𝑒^(𝑖𝑧), like that.
And now, I will keep this as how it is, (𝑒^(𝑖𝑧))²,
move this to the other side, minus 4𝑖,
and then let me also open the parentheses, and this is 𝑒^(𝑖𝑧), and then minus 1; this is equal to 0.
Something squared; this is the same thing up to the first power
This has 0 of that, right?
so this is a quadratic equation in terms of 𝑒^(𝑖𝑧),
so we can use the quadratic formula to solve this.
So, this will tell us the input 𝑒^(𝑖𝑧).
By the quadratic formula,
We know we will have-- the 𝑏-value is this: -4𝑖, then keep that in mind.
So we will have -𝑏, which is the negative right here, and then -4𝑖 inside like that, plus-minus (±),
Square root, 𝑏², 𝑏 is that, right? (-4𝑖)²-4𝑎𝑐
𝑎 is 1; 𝑐 is -1, right?
[Steve Chao gets surprised]
𝑎 is 1, like that.
1 and -1 all of that, all over 2(1).
All right, 𝑒^(𝑖𝑧) equals,
Negative, negative, becomes +4𝑖, this is ± square root; let's do this in our head.
(-4𝑖)², (-4)² is +16, but 𝑖² is -1,
so this is going to give us -16, okay?
and then, (-4)(1)(-1) is going to be +4, plus the -16; we are going to end up with -12, right?
So I can erase this right here for you guys,
and then this is all over 2,
and then this is 𝑒^(𝑖𝑧) equal to, let's break this down.
And let me finish putting 4𝑖 ±, this is the same as saying (4)(3) inside, right?
And then the 4 inside the square root becomes a regular 2 on the outside.
But this was negative,
So we have to take out the 𝑖, and then earlier,
I mentioned the 3, so that's still inside of the radical.
So this is what we have all over 2.
and then you see this is 𝑒^(𝑖𝑧) equal to-- we'll reduce the 2's.
In the meantime, also, let me factor the 𝑖, so let me factor out the 𝑖,
and then 4/2 is 2, 2/2 is 1, and then ± right here;
We have the √(3) like that, okay?
Well, we have 𝑒^(something) equal to something else.
What we usually do?
Of course, we do the typical thing: we can just take the 'ln' on both sides, right?
So that ln and 𝑒 will cancel each other,
and right here, we will just have 𝑖𝑧 equals ln,
and you know this is the product of two things, ln(𝑖) and then inside,
𝑖 times this quantity, so I can break it apart.
This is ln(𝑖), and then plus
because if it's a product, we can just have the sum of 2 ln,
so the second one is ln(2±√(3)), like this.
and if you would like, you would just divide everything by 𝑖, and you're pretty much done.
But let's do this more legitimately, because what in the world is ln(𝑖)?
Just like earlier, we talked about the sine in terms of its complex version,
we should also talk about how can we enter a complex number into the ln function, right?
So now, I'm going to go back to my first board,
and I'll have to talk about the definition of the complex logarithm, real, real quick, right?
So consider you have a complex number, 𝑧=𝑎+𝑏𝑖,
and let's look at the plane, right?
These are the real values and this is the complex axis,
so, let's say we have the complex number right here, 𝑎+𝑏𝑖.
That means from here to here is 𝑎; from here to here is 𝑏, all right? So this is that.
Once again real values and then complex axis, or the imaginary axis.
Anyway, so 𝑎+𝑏𝑖.
Hopefully you guys have seen this, or if you guys haven't, now this is the time.
Another way to look at this--
we can look at it as a polar form, just like the polar equation. It's very similar.
From the origin to here, all right?
Yes, I don't have a color chalk, so yeah.
Anyway,
right here, what I want is I want to know the distance from here to here, I will call that to be 𝑟
and then I need to know what the angle is.
So when you have 𝑧=𝑎+𝑏𝑖,
I can drag this down as,
well, as we know,
just like the complex-- just like the polar equations, we have this right here 𝑎 is equal to the 𝑥 value,
Which is the cosine, right? But you multiply by 𝑟.
𝑟cos(𝜃); likewise, 𝑏=𝑟sin(𝜃),
so if 𝑧=𝑎+𝑏𝑖, 𝑎 is that; I can just put down 𝑟cos(𝜃), plus, 𝑏 is that, so 𝑟sin(𝜃),
and we still have the 𝑖, so let's put the 𝑖 right here, okay?
And now, I can, of course, factor out the 𝑖-- I mean, the 𝑟 and then inside is cos(𝜃)+𝑖sin(𝜃).
And do you guys see the connection?
This right here is the Euler's Formula,
so I can put this into the polar form, just like that, right?
but technically, this is called the polar form.
But anyway, this is 𝑟;
this right here, by the Euler's Formula, is 𝑒^(𝑖𝜃)
because 𝜃, right now, is the input, okay?
And why am I showing you guys this?
Because now, what you have this right here, 𝑧=𝑟𝑒^(𝑖𝜃),
This is just another way to represent the complex number; I can now take the ln on both sides.
So, on the left-hand side, we have ln(𝑧) and this is going to be ln(𝑟*something).
This is a product, so I can break it apart.
ln(𝑟)+ln(𝑒^(𝑖𝜃)), right?
And now, here's the punchline.
ln of that complex number,
Once again, the 𝑧 is the complex number.
Sometimes, it can be real because a real number is also a complex number.
Anyway, ln(𝑧) is equal to ln(𝑟) plus, ln and 𝑒 cancel all so we have just, plus 𝑖𝜃, right?
And at the end, here is a typical thing that we will do in the complex analysis, that kind of thing.
ln of the complex number, this is equal to ln, 𝑟 is the distance from origin to the complex number,
So we make a note.
We use absolute value to denote distance, right?
So 𝑟=|𝑧|.
In this case, just like the typical distance formula, you'll know this is 𝑎²+𝑏².
But instead of 𝑟, I will put down this,
so we will have ln(|𝑧|) to denote the distance from the origin to that complex number, all right?
and then plus, 𝑖 is just 𝑖,
and to be slightly technical this 𝜃 right here is called the argument, and it just put up arg.
Doesn't really matter, I'll just write it down here for you guys at this one time.
But, because I'm not really into complex analysis,
Let me just box this, all right?
𝑟 is the distance; 𝜃 is the angle. That's what I'm going to tell you guys in this video,
but here's the typical thing that we will do.
Anyway, so now,
let's figure out what's ln(𝑖), so let's do that right here.
Question: What is ln(𝑖)?
And now, we have to look at the real axis and then the complex axis, right?
𝑖 is the same as saying 0+1i. 0 is right here, and then 1𝑖 is right here,
so here we have the 𝑖, let me put it down here, all right?
So this is the same as 0+1𝑖.
Okay, so what's the angle from here to here?
𝜋/2, right? So this is going to give angle 𝜃, and what's the distance from here to here?
Just one step up, right?
So we know 𝑟 will be 1, well, |𝑧| will be 1.
So as you can see this is going to be, by that formula, okay? By that formula,
which is going to be ln(𝑟) which is 1, let me denote that, it's supposed to be 𝑟,
which is 1,
and then we add it with 𝑖 times this angle,
I'm going to say this, but I'm not going to write down too much things.
Otherwise, I can never finish this video.
The 𝜃, okay?
It's defined to be from -𝜋 to +𝜋. Do not include the -𝜋, but we do include the +𝜋, all right?
So, it goes like this.
So we are going to use 𝜋/2 from here to here.
Anyway, if you just measure the angle from here to get the usual way, that's +(𝜋/2), all right?
So this is the argument, or the 𝜃, however you want to call it.
ln(1)--- that's a usual ln(1), which is 0.
So, all in all, ln(𝑖)=𝑖(𝜋/2). How cool is this?
So put this down right here.
Now, we're out of space, but it's okay, because I have another one.
All right, so let me just put this down right here real quick.
ln(𝑖)---
Well, I should have to write it down like this.
𝑖𝑧 is equal to ln(𝑖)-- let me just copy down this real quick, ±ln(2±√(3)), right?
So, this is 𝑖𝑧, ln(𝑖), once again, is the number that we got earlier, which is (𝑖𝜋)/2.
So now, this right here is just (𝑖𝜋)/2 and then ±---
Oops, it's just a plus; it's not ±, all right?
Anyway, ln(2±√(3)).
And yes, we have multiple solutions, but anyway,
Now I'm just going to divide everything by 𝑖 so let's check it out, all right?
So I will just put this down like this, multiply by 1/𝑖, so that this and that will cancel,
yeah, you guys know how to do it.
So you have just 𝑧.
This right here, no more 𝑖's, just 𝜋/2, and then plus 1/𝑖 right here, 1/𝑖,
and then this is ln(2±√(3)), okay?
𝑖 in the denominator doesn't make us comfortable,
so let's multiply 𝑖 on the bottom and also 𝑖 on the top,
and you know this is going to be -1 on the bottom,
So we'll change the plus to minus, so all in all, these equal to 𝜋/2 minus,
once again, the minus is because 𝑖² gives us this minus.
On the top, 1*𝑖 is still 𝑖 and then we have this part,
ln(2±√(3)).
Ladies and gentlemen, this is the answer.
One of the answers, well, actually, two of the answers, because we have ± to that equation.
And, perhaps I'll just leave it to you guys as an exercise.
if you are checking out a textbook question and maybe you are working this out, right?
Sometimes, books, they may give you this for this final answer:
we have 𝑧 is equal to 𝜋/2 and they do have ± right here,
and then we still have the 𝑖 and then the ln,
and the inside here is just 2+√(3).
I didn't write anything down incorrectly, okay?
We do have ± right here, and then, inside, it's just going to be 2+√(3), okay?
And to finish this, all together, we can put down plus 2𝜋,
well the integer multiple 2𝜋, because sine is periodic.
So, this may be some of the solutions that you'll see,
but I'll just stop right here.
If there is a request, I'll show you guys how to go on this part to this part.
And once again, the 2𝜋𝑛,
it's just going to be picky to find all the solutions, right?
So, hopefully you guys like this,
I'm going to just put this down and show you guys that again.
sin(𝑧)=2, it does have a solution. In fact, it has a lot of solutions.
You never know what will happen the complex world, so, yeah.
Be careful.