12:19 PM
0

Consider a non-UFD that only has 2 units ( $-1,1$ ) and the min difference between 2 elements is $1$. Also there are only a finite amount of elements for any given fixed norm. ( Maybe that follows from the other 2 conditions ? ) I wonder about counting the irreducible elements bounded by a lower...

Any ideas ??

@mick what does the min difference between elements mean?

12:53 PM
What does the notation $\times_{G}$ mean?

@quallenjäger Depends on context. Here it likely means the orbit space under $G$-action.

@quallenjäger probably fibered product over $G$

I am studying the principal bundle, they write the associate bundle from a principal bundle $P$ as $E\times_{G}F$

@TobiasKildetoft For group actions it's usually orbit space of a product of two G-spaces, in my experience

12:59 PM
In this case it's definitely that though

If a and b are coprime, does that mean that b and a are prime?

observe that because the fibers of $P$ are identified with $G$ as $G$-spaces, the fibers of $P \times_G F$ are identified with $G \times_G F \cong F$, given by sending $(g,f) \mapsto gf$; recall $G$ acts on the fibers on the right
I think I have my right/left handedness all correct there but I occasionally get those wrong

@LeakyNun No, of course not

@Narcissusjewel I sort of didn't respond to your question the other day but I can try now
Ehhhh

@TobiasKildetoft I think he was joking haha

1:03 PM
I'm too lazy to give a broad overview, nevermind
But maybe John Ma will :D

@MikeMiller Haha sure, let me know if you're ever not feeling lazy :p.
Yep I will prod him about it shortly. Have to make coffee or obtain an energy drink.

@MikeMiller What is $P \times_G F$? I am building the cross product over a principal bundle?

@quallenjäger $G$ acts on $P$ on the right and $F$ on the left. We set $$P \times_G F = (P \times F)/(pg,f) \sim (p,gf).$$
Aka, the quotient/orbit space of the product.
this retains a left G-action by acting on the left by G, and this action on each fiber can be identified with the given G-action on F

@MikeMiller I see, thanks

@Narcissusjewel The two-line catchphrase, I'd say, is that it's a type of geometry in which there's a lot more local flexibility than you're used to (all symplectic manifolds are locally isomorphic to a "standard" symplectic manifold, unlike Riemannian manifolds, where curvature is a local invariant)
So you can't possibly tell them apart via simple invariants. Then the features of symplectic situations tend to be global in character, and often feature counts of certain types of (global) submanifolds of symplectic manifolds, or symplectic versions of e.g. the Riemannian volume

1 hour later…
2:13 PM
morning chat
some notes for myself: in spherical trig, one has the half-side angle identities $$\sin\frac{A}{2} = \sqrt{\frac{\sin s\sin (s-a)}{\sin b \sin c}},\; \cos\frac{A}{2} = \sqrt{\frac{\sin (s-b)\sin (s-c)}{\sin b \sin c}},$$
where $a,b,c$ are the side lengths of a spherical triangle, $A$ is the angle opposite to side $a$, and $2s=a+b+c$
hence $$\sin A=2\sin \frac{A}{2}\cos\frac{A}{2}=\frac{2 \sqrt{\sin s\sin(s-a)\sin(s-b)\sin(s-c)}}{\sin b \sin c}$$
and hence $4\sin s\sin (s-a)\sin(s-b)\sin(s-c)=\sin A\sin b\sin c$. cyclic permutation gives two other such formulas.

This part $$\sqrt{\sin s\sin(s-a)\sin(s-b)\sin(s-c)}$$ so reminds of heron's formula

this last conclusion also implies the sine rules $\frac{\sin A}{\sin a}=\frac{\sin B}{\sin b}=\frac{\sin C}{\sin c}$
@Secret agreed.

is that really heron's formula analogue in spherical trig?

I don't know.

2:34 PM
@Secret what I'm really after is a geometric proof / interpretation of the following identity:
$$4\sin s\sin(s-a)\sin(s-b)\sin (s-c)=\det\begin{pmatrix} 1 & \cos a & \cos b \\ \cos a & 1 & \cos c \\ \cos b & \cos c & 1\end{pmatrix}$$

2:46 PM
hey all

oh. error in my above: should be $\sin^2 A \sin^2 b\sin^2 c$

3:03 PM
A cycloid is the locus of a point on the circumference of a rolling circle, and it's arc length for one revolution of the circle is 8R, does this result change if the circle is not moving with constant velocity?
that is the angle doesn't vary with time linearly

@TedShifrin I have the solution.

@MasterYushi it doesn’t as we only consider it rolling

3:35 PM
welp

Hey guys, for a function that is $1$ for positive values of $x$ in $\mathbb{R}$ and $0$ for non-positive $x$, is $f((x_n))_{n \in \mathbb{N}}$ with $x_n = (-1)^n\frac{1}{n}$ divergent or convegent? It alternates between $1$ and $0$, so for every $\epsilon$ there is an $n$ such that for all $N>n$ there is an $N_1$ such that $f(x_{N_1}) - f(x) \ge \epsilon$, so yes it deverges. But $\lim_{n \to \infty}f(x_n) = 0$, so the limit exists, so it conveges. What is right?

3:53 PM
Hello is anyone on?
I have a question thats really bothering me, I hope someone can help me out

@PrittBalagopal Is it along the lines of, what is the meaning of life?
Or, is it more directly related to math?

lol

Math related lol @KarlKronenfeld

@PrittBalagopal yes there some here, but they will answer you only if the question is enough good and they will like your question

Ah, now your question is bothering me, simply because it is shrouded in mystery.

4:01 PM
OhhKayy, so basically, how do you find minimum distance of a point from a surface defined by 3d cartesian equation?

usual method is Lagrangian multiplers

with the distance squared being the objective function and the surface being the constraint.

Darn it, @Semiclassical beat me to it.

(you do distance squared rather than distance b/c that makes the calculation easier without changing the result)

4:04 PM
You do distance square because you can do the calculations in the first place
The distance function is not very differentiable

yeah

@BalarkaSen smoothness is not the issue, because the point is presumably not on the surface

yeah, the distance function will be a smooth function of position so long as the distance never goes to zeor
that said, it's waaay more convenient to minimize distance squared

Oh, man, I got distracted. I was looking something up in a mse question, and then I found my way into the chat!

4:06 PM
@KarlKronenfeld Ah, fair enough.

Okay, I currently don't know what Lagrangian Multipliers are, trying to read a Wikipedia article and see if I can understand anything. Btw, I am just a college first year student, so I know very little math.

4:18 PM
If a series converges uniformly on $[a+\varepsilon,b)$ for all $\varepsilon>0$ and pointwise on $[a,b)$, is then the convergence uniform on $[a,b)$?

@PrittBalagopal the geometric idea goes like this. suppose you've got a surface and a point not on that surface.
if you put a small sphere at that point and expand it slowly, eventually said sphere will touch the surface

Oh, I see

at the moment this happens, the sphere is tangent to the surface and the radius of said sphere is the minimum distsance to the surface

Oh never mind
Yes, and we have to find the radius right?

right
geometrically, what Lagrangian multipliers enforces is that condition that the sphere (i.e. a level set of the objective function) be tangent to the constraint surface

4:22 PM
Oh I see.

algebraically, you do that by demanding that $\nabla F=\lambda \nabla G$ where $F=F(x,y,z)$ is your objective function, $G=G(x,y,z)$ is the constraint function, and $\lambda$ is some nonzero scalar

I looked at the wikipedia page, it says they define a new function L(x,y,z,k), such that L(x,y,z,k)=f(x,y,z)-kg(x,y,z)

the gradients there determine the normal vectors for the tangent planes at the point of mutual tangency, and the above requirement enforces that the two normal vectors are proportional
yeah, same thing. in that case you demand $\nabla L=\nabla f-k \nabla g=0$
I don't think the geometry is nearly so obvious when written like that, though

Oh wait, I don't know what gradients are
Is it like a slope analogy for 3D?

i wondered if you would. it's a vector function consisting of partial derivatives: $\nabla F = (\partial F/\partial x,\partial F/\partial y,\partial F/\partial z)$
it's not quite the slope analogy; that'd be more like assuming a surface of the form $z=f(x,y)$ and taking derivatives $\partial f/\partial x,\partial f/\partial y$
what it does have in common with the slope analogy, though, is that gradients are related to tangents to surfaces

4:27 PM
Question: suppose you had a curve represented by a pair of parametric equations. This curve goes to the same point as t goes to either positive or negative infinity. How would you go about finding a point that is equidistant from all points on that curve in the xy plane?

that's a topic for multivariable calc, though
where this leads to is that, when you use a lagrangian multiplier to optimize an objective function

0

Reference: Linear Partial Differential Equations for Scientists and Engineers(Authors:-Tyn Myint-U, Lokenath Debnath) I understood till the red box. I have doubt on the equation in the red box. How do they take limits of the integral? I don't understand. Please help me. What is the motivatio...

you end up with the following equations: $\frac{\partial F}{\partial x} = \lambda \frac{\partial G}{\partial x}$, $\frac{\partial F}{\partial y} = \lambda \frac{\partial G}{\partial y}$, $\frac{\partial F}{\partial z} = \lambda \frac{\partial G}{\partial z}$, and the constraint equation written as $G(x,y,z)=0$.
as an example, take the surface to be $z=0$ and the point of interest to be at $(0,0,h)$ (i.e. $h$ above the origin.)
the surface constraint is that $G(x,y,z)=z=0$ and the objective function is $F(x,y,z)=(x-0)^2+(y-0)^2+(z-h)^2=x^2+y^2+(z-h)^2$ (i.e. the distance squared between $(0,0,h)$ and a particular point on the surface).

@Semiclassical Is there any way to render the LaTeX commands? I'm not really able to translate all those equations you wrote.

Use the LaTeX in chat link in the room description. that'll give instructions
the above equations then become: $2x=\lambda\cdot 0,$ $2y=\lambda\cdot 0$, $2(z-h)=\lambda\cdot 1$, $z=0$.
so $x=y=z=0$ in which case $F(0,0,0)=h^2$. So the minimum distance from a point at height $h$ to the horizontal plane $z=0$ is just $h$ (as it has to be).
This example is trivial, but the same idea can work for $F,G$ which are decidedly nontrivial.
Any book on multivariable calculus should discuss Lagrangian multipliers, so I'd look there for further examples.

4:39 PM
Wow thanks @Semiclassical ! You told me alot of stuff, I'll take quite a while to fully understand. Thanks for your help, really appreciate it!

So one thing I found intriguing about that positive semidefinite symmetric bilinear form is as follows:
1. Why the diagonal of ones
I always felt like the ones has to have some meaning besides to make the det work

2

In completely unrelated news, given how fast AlphaZero master all chess it seems that soon there will be AI which can learn how to explicitly count to $\omega_1$

I don't understand.

4:58 PM
mathematica...wtf is wrong with you

Stephen Wolfram created me and Im a monster

basically: I've been cutting and pasting a certain displayed MatrixForm object repeatedly
but somehow mathematica's interpretation of that internally isn't the same as what it's showing visually

@KevinDriscoll did you see a new kind of science in the night sky lately?

@Semiclassical Ya Ive seen lots of screwey things happen with MatrixForm
@BalarkaSen uuuummmmm now that I know of?

I see you're not familiar with the meme
Oh well

5:05 PM
ANKS

this man here knows his science

OH BECAUSE OF STEPHEN WOLFRAM

hmm
suppose I've got a positive semidefinite matrix $I+S$ where $S$ is symmetric and zero on the diagonal
then $e_2^T(I+S)e_1=S_{12}$.
Is there an easy way to argue from that that there exist unit vectors $v_1,v_2$ such that $v_2^Tv_1 = S_{12}$?

5:24 PM
Hi
Is the automorphism group of a topological group always topological?

okay, got it: since $I+S$ is positive semidefinite with $S$ symmetric (and real---forgot to say that) there's a cholesky factorization $I+S=V^T V$. Hence $S_{12}=e_2^T (I+S)e_1=e_2^T V^T V e_1=(Ve_2)^T (Ve_1)=v_2^T v_1$.

I felt like there's a whole world of matrix factorisation I did not knew exists until now...
I was only aware of things like QR, Jordan, SVD, LU

yeah

@brot every group is a topological group with the discrete topology. A more interesting question to ask is if there is some topology on the automorphim group which captures some of the topology of the space. If the space is locally compact (or just compactly generated), you can use the compact-open topology

@LeakyNun The unit ball is the same as that in the 1-norm. Thus the metric space might be $L^1(\Bbb{R}^2)$

5:37 PM
@Secret Being a topological vector space, $L^1(\Bbb R^2)$ is homogenous. But $\Bbb R$ with this topology is not homogenous, because, as Leaky observed $\{0\}$ is not open, but the other singletons are

hmm...

7

The post office metric space, $P$ has the distance function defined as follows: $$d_P (\mathbf{x},\mathbf{y}) := \begin{cases} 0 & \mathbf{x} = \mathbf{y}\\ \Vert \mathbf{x}\Vert_2+\Vert \mathbf{y}\Vert_2 & \mathbf{x}\neq \mathbf{y} \end{cases}$$ where $\Vert \mathbf{x}\Vert_2 = \sqrt{x_... hmm, very strange... Wait why are the singletons open? That doesn't sound right Choose any$\varepsilon < |x|$Then the ball around$x$is$\{x\}$5:45 PM Hmm so the topology is almost discrete Since singletons are closed @MatheinBoulomenos Ok, I see. I guess I thought there was some kind of induced topology. @brot the compact-open topology is always defined. But in general composition is not continuous. (It is if you assume locally compact or compactly generated) i need to show that the decomposition of harmonic and analytic function is harmonic, someone can help? i thought that would follow from cauchy riemann equations but it does not seem to work analytic as in complex analytic? I think you can use that every harmonic function is the real part of some holomorphic function and that the composition of analytic functions is analytic so why would it be harmonic? and yes complex analytic 5:57 PM the real part of a holomorphic function is again harmonic (which does indeed follow from Cauchy-Riemann) yea what i just wanted to write :P but if$f$is the analytic function , and$u$is the harmonic one , we have$u = Re(g)$for some analytic function$g$, so you say that$u $composition$f$is the real part of$g$composition$f$, right ?@MatheinBoulomenos right it seems a bit strange to me, you say the$Im(g)$composition$f$is$im(g $composition$f ) $(sorry im not sure what is the sign for the composition) 0 Here is theorem 1.14 from Rudin's RCA: If$f_n : X \to [-\infty, \infty]$is measurable for$n=1,2,3,...$, and$g = \sup_{n \ge 1} f_n$and$h = \lim_{n \to \infty} \sup f_n$, then$g$and$h$are measurable. Immediately following this is the corollary (a): The limit of every pointwis... @Liad the symbol is \circ 6:02 PM thanks @MatheinBoulomenos so you say$im(g) \circ f = im(g\circ f ) $? Hi @Ted hi Mathei a wild @Ted appears LOL, you're quoting Balarka, Eric He's quoting pokemon 6:04 PM @Liad yeah. Think about these as functions$\Bbb R^2 \to \Bbb R^2$. Taking the imaginary part is just projecting on the second coordinate Here's one for you, Eric. Preliminary analysis of Leaky's post office type metric @MatheinBoulomenos sounds right, i'll try to prove it, thanks! I gave up on trying to use Hatcher to do the algebraic curve thing @Ted settled for invoking implicit and doing that oh this is interesting Eric: I think my suggestion of Riemann-Hurwitz is the fastest way using just alg top. 6:06 PM The unit balls change shape with their centres. With centres$|x| > 1$no unit balls exists Plotted there are the balls of radius 1,2,3 for 3 different centres, colored coded Do you guys get something similar? @TedShifrin It's just lifting the orientation from the base using the local biholomorphism, yes? yeah I was trying to go down that direction but my brain exploded cause i do way too much math (analysis will continue tomorrow as once again it is getting late) (biholomorphism is an orientation preserving diffeomorphism) I'm lost, Balarka. You were suggesting classification of surfaces. Because of the branch points, you fail to have a local biholomorphism everywhere. @TedShifrin Yes, but Eric seems to only want to prove orientability (or computing$H_2$, equivalently) You can do that by ignoring the branching locus I'm guessing because he's saying "settled for invoking implicit and doing that" i was only going for orientability Right The point is Ted's technique gives you$H_1$too What's the proof that orientability still follows even if you've pulled out finitely many points? Don't start telling me$w_1$stuff. the complex analysis/metric problem you just linked is p good @Ted 6:16 PM Can't you use the generalization of Liouville's theorem from Krantz? Hi @Ted Hi @Alessandro I don't have my resources any more to look up, Mathei. @Maneesh: That's a long post. Can you ask a specific question? It looks like it follows immediately by substitution, as the authors say. @TedShifrin I was going to tell you that non-orientability implies there is an embedded circle in your manifold which has nontrivial normal bundle. But I guess that falls into the category of "$w_1$stuff" I have an answer but im too lazy/busy doing other work to write one up so gg Yeah, so you need to push that circle off the missing points. 6:18 PM Right I guess I can't see an elementary enough proof Let$U$be an open subset of$\Bbb C$equipped with a metric$\sigma$such that the curvature$\kappa_\sigma$satisfies$\kappa_{\sigma}(z) \leq -B <0$for some positive constant$B$and all$z \in \Bbb C$, then any holomorphic function$f: \Bbb C \to U$is constant And so you take the map$f(z)=e^z$? @Mathei This is in Krantz "Complex Analysis - The Geometric Viewpoint" p 73 @MatheinBoulomenos yeah that @TedShifrin yeah 6:20 PM Complex analysis: A geometric approach but like actually I think there's a direct way to get it from the Ahlfors Lemma, as the OP suggested. @EricSilva he didn't want to get sued by Ted @TedShifrin I guess you can use compactly supported cohomology too. you can prove that fact from Ahlfors @Ted easily enough That's getting fancier, @Balarka. 6:21 PM Somewhat Easier to construct a nowhere-0 2-form in the first place, namely, the restriction of the Kähler form on$\Bbb P^2$. I was going to comment that that doesn't work for noncompact manifolds until you added the last bit I agree only one final remaining Yippee. @TedShifrin Why do they integrate$\frac{f(\xi)y_1(\xi)}{p(\xi)W(\xi)}$in the interval$[a,x]$? and$\frac{f(\xi)y_2(\xi)}{p(\xi)W(\xi)}$in$[x,b]$6:25 PM Maybe i wont leaf through Bryant's EDS this break and define green function like that? Yes, that is suspicious, @Maneesh. They seem to have the wrong sign on the derivative of$u_1$, since$x<b$. But it works out right when they put the negative sign in the second line. Then you get two negatives when you use the fundamental theorem of calculus. So the first line is typical physicist nonsense, but they get the right answer. @TedShifrin sir, i don't get the point. I understood, Why there is negative sign. So you're asking why they use$\int_a^x$for one and$\int_x^b$for the other? Because that actually gives the Green's function formula they want. The Green's function should depend on the entire interval. ok sir. Now I understood. they have used the fundamental theorem of calculus to denote$u_1(x)$and$u_2(x)$right? 6:39 PM > physicist nonsense Yes. Thank you very much sir:) Sure thing. im trying to show that$v(z) = 1/(2\pi) \int_0 ^{2\pi} Re(\dfrac{e \ ^ {it}+z }{e \ ^ {it }-z}) u(e \ ^ {it})dt$equals the real part of some analytic function.$u$is harmonic. Someone can help? "The end justifies the means." re "nonsense" :P 6:41 PM Ah :P those physicists know a surprising amount about what's going on despite writing so much crazy nonsense Hey there folks! Hi @Daminark run for the hills 6:42 PM Looks a lot like Poisson's formula, @Liad. yea @TedShifrin it is Oh, indeed, Eric. But I said that for Kevin and Semiclassic :) disappears into the hills the$Re(..) = P(z,e \ ^ {it})$So write the line integral without the real part as a line integral over the circle, @Liad. What do you have? How's it going? (You were wise to run because now I... I didn't have an evil plan prepared, go ahead) 6:44 PM This is the same sort of thing you've been having repeatedly ... where we took the derivative under the integral sign months ago. Ah, Kevin is psychic. @TedShifrin you mean write this:$1/(2\pi) \int u(e \ ^ {it} ) dt$as a line integral? Hi @Daminark Yo No, you need the other stuff in there with$z$and$\zeta = e^{it}$. We're finally doing some nontrivial stuff on Riemann surfaces 6:46 PM Like what? Poisson is one of my 9/10 formulas Poison is great 2$Im(Poison) = Re(Death)$@TedShifrin what to do with$u$? A-hah! @TedShifrin is here. 6:47 PM$u$stays ... it's a continuous function. heya @Antonios. I saw your ping a while ago. @Mr.Xcoder: That's totally fishy. The result in the problem follows rather neatly from JNF :) @Ted Proving that$M(\Bbb C/ \Lambda)= \Bbb{C}(\mathscr{P},\mathscr{P}')\mathscr{P}$is the best I could do for Weierstraß P @TedShifrin ¯\_(ツ)_/¯ moduli space of torus ? Aha ... Use mathfrak, Mathei :P 6:48 PM oh, field of meromorphic functions @TedShifrin You know what they say. One man's fish is another man's poisson. 2 blah So you get the plane cubic, too, Mathei.$\mathfrak{P}$I use$\wp$6:48 PM yeah, \wp Sorry im not sure what you mean, if i leave$u$i cant use Chauchy's formula @TobiasKildetoft Be sure to order the poisson rather than the poison ;) my fault ... not mathfrak. thanks @Balarka (basically stolen from irregularwebcomic.net/974.html) 6:49 PM can anyone here actually physically write in mathfrak tho @TedShifrin not sure what you mean =@Liad: why not? I had to when I taught Lie algebras, Eric :P @MatheinBoulomenos Any plane cubic is birational to a torus smooth plane cubic in CP^2 You get the embedding of$\Bbb C/\Lambda$as the projective plane cubic. @EricSilva I can do decent$\mathfrak{p}$'s and$\mathfrak{q}$'s. 6:50 PM i can't do mathfrak to save my life actually just biholomorphic I think @Antonios: I believe JCF, but what is the right statement to prove? Remember that I suggested certain eigenspaces got combined for$p(A)$. takes away Eric's life is slain hm , we got something like$1/(2\pi) \int \dfrac{e \ ^ {it} +z }{e \ ^ {it} -z }u(e \ ^ {it}) dt$, dont we? yes, Balarka, biholomorphic. 6:51 PM I always suspected that the mathfrak font would be somehow related to my demise Rewrite$\zeta = e^{it}$and write the line integral. @TedShifrin It follows after you reparameterize in CP^2 to bring it down to the Weierstrass form IIRC You don't have to do that, @Balarka. It follows from facts about the doubly-periodic function$\wp$. Well, true. I think pictorially I guess @TedShifrin: Basically, after some thought, the claim is : If the characteristic polynomial of$A$is$p(x)=\prod_{j=1}^k (x-\lambda_j)^{d_j}$, then the characteristic polynomial of$q(A)$($q$some polynomial) is given by$\prod_{j=1}^k (x-q(\lambda_j))^{d_j}$. 6:53 PM I'm not sure what you mean by Weierstrass form. Why is that "pictorial"?$A$is over$\mathbf{C}$, so that using JNF$J$for$A$, and taking$p(J)$the result is immediate. OK, Antonios. I believe that. You still haven't told me "the result." Oh the claim is that the characteristic polynomials correspond like I stated above But that doesn't tell me the eigenspaces. (replace$\lambda_j$with$q(\lambda_j)$.) Sorry if that wasn't clear. 6:54 PM we get$\dfrac{1}{2 \pi i } \int_{|w| = 1} \dfrac{u(w)}{w-z} + \dfrac{z}{w(w-z)} dw$? @TedShifrin Ugh, @Liad. That doesn't look right. agree :P Spare me the partial fractions. What is the formula? I already showed that the eigenspaces are the same @TedShifrin. But, that also follows from the JNF. But I claim they're NOT the same, @Antonios. 6:55 PM Oh, huh. Why doesn't the JNF show that? You may have to combine eigenspaces of$A$to get eigenspaces of$q(A)$. That was my whole point.$f(w) = 1/(2\pi i ) \int_{\gamma} \dfrac{f(z)}{z-w} dz$@TedShifrin Huh? @Liad This is what you get when you rewrite, removing the real part? Um, no. no i thought you asked what's Chauchy's formula I'm not asking for Cauchy's integral formula. 6:57 PM @TedShifrin$y^2 = x^3 + ax + b$Oh, right, @Balarka: But we have to use the facts about zeroes/poles of$\wp$to get such a thing. I don't see that that is pictorial. I think it's kind of pictorial because you can choose a slice of$\Bbb C^2$by a 1-dimensional subspace so that it visually has two circle components (with one point on one of the circles at infinity, so that it's parabolic). @TedShifrin Hm, really? We're discussing totally different things. I thought it was just a parameterization trick$1/(2\pi) \int_0 ^{2\pi} \dfrac{e \ ^ {it}+z }{e \ ^ {it }-z} u(e \ ^ {it})dt\$ @TedShifrin this is what i get from removing the real part, what did you want me to do from here?