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Complex trigonometric functions

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So if the real-valued trigonometric functions correspond to the (unit) circle, then what do the complex trigonometric functions correspond to? Those wonky surfaces living in 4D space?
>>
>>8121778
Obviously the unit 4-sphere.
>>
They correspond to the shape that satisfies f''+f=0
>>
>>8121778
The complex "unit circle" is the set of solutions to [math]x^2+y^2=1[/math] for complex [math]x[/math] and [math]y[/math]. This object is a 1-complex-dimensional (2-real-dimensional) manifold which contains both circles and hyperbolas a cross-sections. It is unbounded.
>>
>>8121796
Why?

>>8121789
Wrong, that would treat the real and imaginary parts of a complex number the same way.
>>
>>8121798
Is the manifold embedded in 4-dimensional space? Does it have a common term for it? Any pictures?
>>
>>8121778
cos(ix)=cosh(x)
i*sin(ix)=sinh(x)
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>>8121798
>The complex "unit circle"
The unit circle is already in the complex plane.
>>
>>8121951
Why is it that going complex makes everything hyperbolic?
https://m.youtube.com/watch?v=FfiQBvcdFG0
All of these surfaces look somehow hyperbolic, are they hyperbolic?
>>
>>8121996
>Why is it that going complex makes everything hyperbolic?

Because a circle with an imaginary radius is a hyperbola.

x^2+(iy)^2=(ir)^2 => r^2 = y^2 - x^2
>>
>>8121778
> what do the complex trigonometric functions correspond to?

e^(i*z)=cos(z)+i*sin(z)
cos(z)=(e^(i*z)+e^(-i*z))/2
sin(z)=(e^(i*z)-e^(-i*z))/2

The real trigonometric functions are just the real and imaginary components of e^(i*z) for real z.

e^z converts log-polar coordinates to Cartesian coordinates. e^(a+b*i)=e^a*e^(b*i) = e^a*(cos(b)+i*sin(b)). I.e. ln(|e^z|) is Re(z) and arg(e^z) is Im(z).
>>
>>8121996
Does it give any additional insight whatsoever to inspect the actual geometry of complex functions in 4-dimensional space? I see this isn't usually done?
Is the geometric aspect of complex functions somehow trivial?
>>
>>8121798
please dont call that a unit circle, because it is not a circle.

it's some algebraic variety, maybe of interest, but let me tell you what it is not:
a circle
>>
Can someone please post a picture representing the manifold that the complex trigonometric functions correspond to?
>>
>>8121798

Interesting! If x and y are complex, such that x^2 + y^2 = 1, is there a z such that x=cos(z) and y=sin(z)? Is z unique?

How do you prove that?
>>
>>8122470

If you multiply x^2+y^2=1 out, where x = a + i b and y = c + i d, you get

a^2-b^2+c^2-d^2=1
and
a b+ c d = 0
tough to graph
>>
Can someone please render a projection of the Riemann zeta function rotating in 4d space?
Surely this is simplistic for some of you to implement in Mathematica?
>>
>>8121778
The problem with your question is the logic.

You think that trig functions correspond to circles so complex trig functions must correspond to something equally fundamental. They don't.
>>
>>8122567

Suppose
x = cos(z) = (exp(i z ) + exp( -i z))/(2 )
y = sin(z) = (exp(i z ) - exp( -i z))/(2 i ). Then
x+i y = exp( i z). Let
z = u + i v, so
exp(iz)=exp(-v) exp(iu), and
|x+iy|=|exp( i z)| = exp(-v). Then
v=-log(|x+iy|) and
exp(i u) = (x+iy)/|x+iy|. Since the right
hand side is a point on the unit circle,
there is a u, unique up to the addition of 2 pi k
for integer k that satisfies this.

You could also use
x-i y = exp( - i z) = exp(v) exp(-i u), but you get the same answer, because x^2+y^2=1 implies
(x+i y)=1/(x-i y).
>>
>>8122749
There is no logical problem with my question. I didn't assume the complex geometry to be fundamental like the circle is, I just wanted to know what it is. Now that I know, I think it's interesting and would like to know more about it despite it not being "fundamental".
>>
Does complex (algebraic) geometry deal a lot with the actual geometry of complex functions as surfaces in 4d space?
>>
>>8123661
I've been interested in this too and I havenĀ“t found good information on it yet. What I would like is to have some kind of "polar coordinate system" for [math]\mathbb{C}^2[/math] which specifies points by giving a point on [math]x^2+y^2=1[/math], which would have until length in an appropriate sense (which may not be an actual metric) along with a complex-valued direction. Then the trigonometric functions would behave as they are supposed to when looking at triangles, etc., with complex-valued side lengths.
>>
>>8124019
>complex-valued direction

I meant length here, the point on [math]x^2+y^2=1[/math] is the direction.
>>
>>8122573
So does this surface have real dimension 2? Four variables a,b,c and d, and two constraints
a^2-b^2+c^2-d^2=1
and
a b+ c d = 0.

Is there a nice parameterization of the surface by the plane, i.e. C itself, or some nice subset thereof?
>>
Isn't this a fairly interesting thread?
>>
>>8124007
no, geometry stops at 3 dimensions.
>>
>>8125209
Does it?
>>
>>8124007
Complex polynomials, yes.

Affine varieties are literally just the zero locus of some of polynomials. i.e. Take [math]\left\{ {{f_1},..,{f_k}} \right\} \subseteq \mathbb{C}\left[ {{x_1},...,{x_n}} \right][/math] then you can define the affine variety [math]V\left( {{f_1},..,{f_k}} \right) = \left\{ {x \in {\mathbb{C}^n}|{f_1}\left( x \right) = ... = {f_k}\left( x \right) = 0} \right\}[/math].

Even more general algebraic varieties still must look like the zero locus of a set of polynomials locally.
>>
>>8126543
Well, what field would generally deal with the geometry of complex functions as surfaces in 4-dimensional space?
>>
>>8126580
>what field would generally deal with the geometry of complex functions as surfaces in 4-dimensional space?
Surely this question is simplistic to answer in terms of what field of mathematics deals most with the actual geometry of complex functions as they are defined?
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>>8127017
We don't just study the graphs of arbitrary functions in geometry. Because the ""surfaces"" these graphs create are not always well behaved.

The behavior of these complex trigonometric functions is just a topic of complex analysis.
>>
>>8127116
>We don't just study the graphs of arbitrary functions in geometry. Because the ""surfaces"" these graphs create are not always well behaved.
There should be a geometric description for the surfaces corresponding to non-well-behaving functions, shouldn't there, at least when the behaviour isn't pathological over some sensible limit, no? Surely the geometric description of such surfaces is far from impossible?
>>
>>8127134
The thing is they aren't always actually surfaces (from either the differential or algebraic point of view).
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>>8127158
What are they in those cases?
>>
>>8127158
Are they not surfaces because they aren't manifolds? What are they?
>>
>>8127233
>>8128646
I'm sure such objects are defined somehow in mathematics?
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>>8124252
>Is there a nice parameterization of the surface by the plane, i.e. C itself, or some nice subset thereof?

z |---> (cos(z),sin(z)) obviously.
>>
>>8129278
Sometimes things don't have nice structure. I don't know what you want to hear.
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>>8129717
What I want to hear is what such not-quite-surfaces can be defined as. What is a suitable generalization that allows them to be defined? They might not have nice structure but they don't have an impossible structure either.
>>
>>8129420
Is that really it, though?
>>
>>8130733

Yes. You have to limit it to a strip -pi=x<pi to get a bijection, because cos() and sin() are 2pi periodic, but that's it.

>>8122887
>>8124252

If a and b are complex, with a^2+b^2=1, then there is a unique z in that strip such that a=cos(z) and b=sin(z). You find that z=x+iy by setting y= - log(|a+i b|) and exp(i x)=(a+ib)/|a+ib|.
>>
>>8121778
Along the imaginary line they turn into their hyperbolical counterparts cos(bi) = cosh(b) and sin(bi) = sinh(b). Other than that I don't know how to interpret them intuitively.
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>>8129740
>What is a suitable generalization that allows them to be defined?
"topological space"
You're welcome.
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>>8129420
what's the metric and curvatures?
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>>8133003

z=x+i y

F=[
Re(cos(z))
Im(cos(z))
Re(sin(z))
Im(sin(z))]

F = [
cos(x) cosh(y)
-sin(x) sinh(y)
sin(x) cosh(y)
cos(x) sinh(y) ]

dF = [
-sin(x) cosh(y), cos(x) sinh(y)
-cos(x) sinh(y), -sin(x) cosh(y)
cos(x) cosh(y), sin(x) sinh(y)
-sin(x) sinh(y), cos(x) cosh(y)]

The metric is conformal!

dF'*dF = cosh(2 y) * I

where I is the 2x2 identity matrix.
>>
>>8133003
>>8133058

Conformal metric:
p(x,y) = (cosh(y))^(1/2)
Gaussian curvature:
K = -(\Delta log p)/p^2

log p(x,y) = 1/2 log cosh(y)
d^2/dy^2 log p(x,y) = 1/2 cosh^(-2)
p(x,y)^2 = cosh(y)

K = -1/2 (cosh(y))^(-3)

Does this mean that for large y the surface gets nearly developable?
Thread posts: 46
Thread images: 1


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