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.

-ln(2+-sqrt(3)), https://www.youtube.com/watch?v=Me9yxROZbRQ

*Sorry I forgot the square root. |z| =sqrt(a^2+b^2)

**Also, I should have written the horizontal axis as "Re" and the vertical axis as "Im"

This definition of sin(x) in complex terms remember a lot the hyperbolic sine

in the Euler's eqn theta is represented by Z and we solved for that Z, but what my question is that how do you represent a complex value for theta?

Guys i might be missing something obvious but could someone tell me why the e^-iz = cos(z) – sin(z) , i cant figure it out for the life of me i think im having a brain fart

M indian

Mistakely come here😂

(1+4n)pi/2+ln(sqrt(7+-4sqrt(3))) contains infinitely many solutions (I think)

(for n in Z) (or maybe n in N)

Wait, that's illegal.

Ohh myyyy gods ,that was amazing , i had a nerdgasm

OKAY THIS VIDEO IS SO SATISFYING SO CALMING, IDK, I JUST GOT A MATHGASM RN 😂😂 I'VE BEEN AWAKE SINCE 18 HOURS, AND I AM THINKING ABOUT WATCHING ANOTHER BPRP VIDEO😂

AM I THE ONLY ONE WHO NOTICED THE DERIVATION OF FIBONACCI'S GENERAL TERM AT 16:43? (In the erased part I mean 😂)

When i make video spees x2, your speaking becomes as if you speak chinese

I’ve never seen such a cool math video before ..

Why does no one use parenthesis?

From context I do know what you mean by r cosq + r cosine closes without parenthesis.

r cosq + r sinq could be any of the following:

r cos(q) + r sin(q)

r cos(q + r) sin(q)

r cos(q + r sin(q))

To show that cos(-z) = cos (z) for complex z 😂. From the definition of cos (z) = {e^(iz) + e^(-iz)}/2, the proof is trivial.

How is pi/2 + – i ln(2+sqrt3) = pi/2 – i ln(2 + – sqrt3) ?

arcsin 2

i^i = Exp(-pi/2) – very interesting, because is real number!

I was always wondered about sin z = 2

iz=ln(i)+ln(2±√3)

If I multiply 4 both sides,

[4 ln(i)=ln(1)=0]

So,

z=-i ln(2±√3)

😶

Thank you

z=sin2 where 2 is in radians

Madarchod pagal kuchh bhi kr raha h…sala jaoaani🤣🤣

Theirs a problem in the vetagors …. |z|=√a^2+b^2

Also you r awesome ,thank you for sharing the information ❤️