Math for fun, sin(z)=2

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.

About the author


  1. -ln(2+-sqrt(3)),

    *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"

  2. 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?

  3. 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



  6. 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))

  7. 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.

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