Mathematics - 2016-10-19 (page 1 of 2)

Ted Shifrin

00:02

@dsillman: You've transmogrified into a conic!

dsillman2000

Lol yeah I found this picture and just decided to use it :)

Ted Shifrin

Do we know exactly what's going on?

@dsillman: Did you settle the argument with 0celo? If I know $x>2$ and I say $x>1$, it's not incorrect. It's just not the strongest possible statement. These things happen.

dsillman2000

Yeah I know we settled it. I was just upset because the bounds weren't perfect

Ted Shifrin

It doesn't mean that $x$ is forced to be chosen in $(1,2]$.

We say an inequality (or bound) is sharp if it is best possible in the sense you're worrying about.

0celo7

Yeah I should have mentioned that.

Ted Shifrin

00:10

It's like choosing the supremum (or least upper bound) instead of a random upper bound.

You'll get to that in Spivak :P

0celo7

Spivak?

There's a billion Spivaks

At least 8

Ted Shifrin

I recommended Spivak's Calculus ... Only 8 if you're counting the 5 volumes of diff geo separately.

0celo7

I am

Ted Shifrin

Then I get to 8, counting Calculus, Calculus on Manifolds, and his new Physics book.

0celo7

Yep

Ted Shifrin

00:12

No one counts them as 5 separate books, but OK.

Now, you see, when you say a billion and you mean 8, that's an invalid inequality.

0celo7

@TedShifrin b and 8 look similar.

Ted Shifrin

Sure they do, cuz neither is an embedded submanifold.

0celo7

what?

they don't have the same homotopy type

Ted Shifrin

That, too.

0celo7

@TedShifrin If a family of polynomials is uniformly bounded on a compact interval, can I also uniformly bound the derivatives?

Oh, and the degrees never exceed $k\in\Bbb N$

Ted Shifrin

00:21

Good thing you added that last condition.

Does a $C^0$ uniform bound imply a uniform bound on all the coefficients?

0celo7

That's what it boils down to

Ted Shifrin

Of course, this sort of thing is easy if you complexify and use Cauchy Integral Formula.

0celo7

Oh?

Ted Shifrin

Yup, there's an integral formula for the derivative.

0celo7

I know

But why is that significant here?

Ted Shifrin

00:23

So it gives you uniform bounds on the derivative on compact subsets strictly contained inside your curve.

I dunno. I'm thinking about something else now.

Mike Miller

morning

Ted Shifrin

00:39

G'night.

Mike Miller

00:53

I just walked next to UC admins for about 10 minutes. At one point one of them complained about the fact that people had told them "I'm being asked to teach sections with 50 people!" and laughed it off as "Yes, that's what budget cuts are, sweetie."

After that they talked about how they flew out to Sacramento to eat lunch with the CFO of best buy.

user228700

02:01

Hi everyone :-)

user228700

When given a single point on a line, how to find the equation of the line..? Is this possible?

arctic tern

@Kaumudi you know there are many different lines through a given point right? do you mean a line through the origin and another point?

user228700

@arctictern Yes, I do know that, so alright, there is no way :/

user228700

What I'm trying to do is find the parametric form of a tangent to a circle.

user228700

The equation of a circle in parametric form is given by $x=acos\theta$ and $y=asin\thete$ where $a$ is the radius of the circle.

user228700

02:07

And I've to find the general equation of a tangent to this circle.

user228700

I know that the tangent will intersect the circle at some point $(acos\alpha, asin\alpha)$...but I dunno anything else about the tangent :/

user228700

Oh hang on, I know that the slope of this tangent would be $(-1/m)$ where $m$ is the slope of the radius connecting the point of intersection and the center...

Secret

yup, with the intersection and the slope of the radius, you have the slope of the tangent, hence you have its point slope form and thus can convert it into parametric form

user228700

Yes, but I'm not arriving at the correct expression :/ I keep getting $xsin\alpha+ycos\alpha=a$ The $cos\alpha$ and $sin\alpha$ should be swapped :/

user228700

My brain is effed up in the morning time. I got it. Thanks :-)

arctic tern

02:26

the line tangent to the circle of radius r at (a,b) will be ax+by=r^2

if you're familiar with vectors there's a good argument

Secret

[Division by zero] I have been summarising my findings on that based on an algebraic structure I have been working on in the last 4 days. It is now summarised in a blog article of mine

Arctic, if you don't mind, could you have a look at it and see if it makes sense to you?

Abstract: (The process of building) division by zero algebra seems to provide useful tools in studying generalised RPS magmas

arctic tern

Arctic

The Arctic (/ˈɑːrktɪk/ or /ˈɑːrtɪk/) is a polar region located at the northernmost part of Earth. The Arctic consists of the Arctic Ocean, adjacent seas, and parts of Alaska (United States), Canada, Finland, Greenland (Denmark), Iceland, Norway, Russia, and Sweden. Land within the Arctic region has seasonally varying snow and ice cover, with predominantly treeless permafrost-containing tundra. Arctic seas contain seasonal sea ice in many places. The Arctic region is a unique area among Earth's ecosystems. For example, the cultures in the region and the Arctic indigenous peoples have adapted to...

themoreyouknow

maybe I'll look at it sometime. division by zero doesn't interest me much tbh.

Secret

ok

Brody

04:20

@MikeMiller That's pretty sad tbh

Prince M

05:44

Question: In the proof found here proofwiki.org/wiki/…

When they instantiate the existence of x in the preimage... how do they know such an x exists

arctic tern

@PrinceM the whole point is to conclude the preimage of an open set is open. case 1: the preimage is empty, so it's open. case 2: the preimage is not empty, so we can let x be in the preimage [etc]

Prince M

06:04

yeah, thats what I wrote when I proved it.

My question wasn't exactly what happens if U is empty, it was 'is there a reason why we can assume its not?'

arctic tern

because if it's already empty the preimage is already open

Prince M

So then their proof should say 'let U be a nonempty open set in M2'

iwriteonbananas

07:59

I find this question silly: math.stackexchange.com/questions/1975019/…

Soham Chowdhury

10:13

@BalarkaSen I ... uh ... no, I guess.

also, wow, how are there all these new posts on neverendingbooks

which one do you mean

Balarka Sen

@SohamChowdhury let me look

Soham Chowdhury

Peter Scholze stuff yay

Balarka Sen

neverendingbooks.org/the-oracle-on-the-symmetries-of-roots

@iwriteonbananas It seems like he's asking why should we define the cup product like that. That's perfectly fine?

user116211

If suprema and infima exist, aren't they unique?

user116211

I was reading Kelley, where he writes:

user116211

10:27

> Clearly suprema and infima are unique in chains.

user116211

Isn't it a trivial fact to mention?

user116211

hmm.

Soham Chowdhury

@BalarkaSen oh, this is really interesting

I suppose $\mathrm{Gal}(\Bbb Q^{\mathrm{cyc}}/\Bbb Q)$ is described by CFT?

Balarka Sen

Yea

Soham Chowdhury

this is so cool though

thanks

Balarka Sen

10:31

I thought it was cool too.

I don't know anything about it though

Soham Chowdhury

i just read the punchline

whoa

lol

"Mesdemoiselles Adele et Idele"

Balarka Sen

heh

user116211

Original Bourbaki?

iwriteonbananas

@BalarkaSen But like...the answer is just contravariance

diagonal induces a map on cohomology that goes the other way

so we get a product

Balarka Sen

@iwriteonbananas Sure; but why does this product geometrically mean what it should?

user116211

10:37

It's the French Edition.

Balarka Sen

I think that's what he's asking.

The algebra is clear, but not the motivation.

iwriteonbananas

I dunno, I still find the question off-putting

We get a graded algebra structure on cohomology of spaces

Like....what type of answer is he expecting?

Balarka Sen

The corresponding stupid map for homology also gives $H_*$ a graded algebra structure.

But that ain't interesting

iwriteonbananas

Stupid map = induced by projection?

Balarka Sen

yeah

iwriteonbananas

10:43

Well because we discard whatever is happening on the other factor

The diagonal doesn't discard anything

Lozansky

How can you see that $$p.v. \int_{-\infty}^{\infty} \frac{cos x}{x+i} dx \neq Re \: p.v. \int_{-\infty}^{\infty} \frac{e^{ix}}{x+i} dx $$

iwriteonbananas

I don't know why I'm getting so worked up over this

Balarka Sen

sure, I'm just saying "we get a graded algebra structure on cohomology" is not sufficient reason to be interesting.

@iwriteonbananas if that question puts you off, fuhget about it.

I don't care enough to post an answer there, for example

Lozansky

Is it because of the stupid $i$ in the denominator?

Balarka Sen

It's more of a meta question.

iwriteonbananas

10:46

@BalarkaSen That's true I suppose

Balarka Sen

so, what're you upto?

iwriteonbananas

Studying linear algebra

you?

Balarka Sen

nothing as of now; but studying vector bundles and diffgeo in general

iwriteonbananas

That's good stuff

Too bad I forgot it all

You can re-teach me

Balarka Sen

I'm learning differential geometry of surfaces, though, not the abstract stuff.

But I can teach you whatever I have learnt of vector bundles, sure.

iwriteonbananas

10:55

Sweet

Balarka Sen

11:10

"For God is the Alexandrov compactification of the universe"

-Groth. IV. 22

s.harp

what if the universe already was compact? does god have nothing to do with it anymore?

Balarka Sen

That's what atheists claim, probably. But in truth, it's only locally compact ;)

user227867

12:03

Are there any good sources for transcendental methods in algebraic geometry other than Griffiths and Harris's Principles of Algebraic Geometry? It seems that book is full of errors, according to reviews I read. Thanks.

Balarka Sen

What does "transcendental methods" mean?

user227867

Using the methods of complex analysis and differential geometry, not just algebraic methods like schemes, which is done in Hartshore. Refers to algebraic geometry done over the complex numbers, of course.

user227867

I know about the recent book by Donu Arapura, Algebraic Geometry Over the Complex Numbers, which seems to be a good option too.

Balarka Sen

For curves I have heard praises of Forster. But I don't know any other book on complex geometry. Ask Ted.

user227867

I just bought a copy of Forster's Lectures on Riemann Surfaces. Thanks.

user227867

12:06

Together with it, I got a copy of Gunning and Rossi's Analytic Functions of Several Complex Variables.

user227867

I am also considering getting Huybrecht's Complex Geometry, but it has more than 15 unproved theorems in the book, which is a minus point. However, he does prove the Kodaira vanishing and embedding theorems in full.

Balarka Sen

I have heard of Demailly's book.

user227867

Yes, but it is not yet published, only available as a free PDF on his website

Balarka Sen

ah, k

user227867

I also know about Well's Differential Analysis on COmplex Manifolds and Morrow and Kodaira's Complex Manifolds.

user227867

12:09

Hard to make a choice for me because I have idiosyncratic requirements for books

Ramanujan

12:55

@DHMO

DHMO

@Ramanujan

Ramanujan

Typing question

Area of triangle = 1/2

From were do we get $c_3$ ? As 1 1 1 ?

DHMO

@Ramanujan $c_3$??

Ramanujan

Yes

DHMO

sorry, no idea

Ramanujan

13:00

OMG!!!

DHMO

you could always expand it to verify it

Ramanujan

Are we just adding zero matrix? uploading my work

DHMO

@Ramanujan no, matrix determinants don't work like that

Ramanujan

OK another question to clear

Iam not getting any way to go beyond this step

DHMO

use the formula?

Ramanujan

13:07

DHMO

@Ramanujan try to expand 2)

Ramanujan

I didn't get what you mean?

DHMO

expand the second option

Ramanujan

Iam not interested to solve this questions by verification method , I want to solve given condition to get any one of these options .

Which I tried but stuck at that step

DHMO

@Ramanujan too bad then

Ramanujan

13:13

Why?

Isn't there any way to get that??!?!?!??

DHMO

I advise you to expand from the options, lol

Ramanujan

Ah,didnot expected from mathematician like you😔 you discouraged me,😥

DHMO

@Ramanujan mathematics is discouraging

Ramanujan

You as a mathematician is discouraging me (never ever mathematics) , I will try that on my own

DHMO

@Ramanujan yep, I am discouraging

Ramanujan

13:18

@DHMO you might need some sleep(if not so)

DHMO

@Ramanujan make it one fraction

@Ramanujan why i told you to expand from the options is that there is no "standard" form. your answer is correct on its own

Ramanujan

Mean?

DHMO

to make it as an MC (multiple-choice question) is a kind of conformity

to expand from the options is just another kind of conformity

@Ramanujan fraction addition

Ramanujan

I will be keeping it in my mind

DHMO

of course real world mathematics does not need you to conform into one of the forms

so you don't need to expand from the "options"

because there is no such thing as "option" in real world mathematics

Ramanujan

13:21

Fraction addition?what step you want me to do as? You are right (there is no option in math)

user116211

@Ramanujan $$\begin{bmatrix}x_1 & y_1 &1\\ x_2-x_1 & y_2-y_1 & 0\\ x_3-x_1 & y_3 -y_1 & 0\end{bmatrix}$$

user116211

Now, find $\rm det$ along $\rm c_3\,.$

DHMO

$\displaystyle = \frac12 \left[ \frac{(a-c)ac}{abc} + \frac{(c-b)cb}{abc} + \frac{(b-a)ba}{abc} \right]$

Ramanujan

@MAFIA36790 do you mean this ^?

DHMO

13:33

@Ramanujan did you read my message?

Ramanujan

Can't understand😑 @DHMO

DHMO

@Ramanujan do you know how to add two fractions?

Ramanujan

Yes

DHMO

three fractions?

Ramanujan

Lol to read that “do you know how to add two fractions?”

DHMO

13:36

@Ramanujan then just add the 3 fractions tgt

Ramanujan

Yes I can

Do you mean 1/a + 1/b + 1/c ?

DHMO

no

@Ramanujan here

34 mins ago, by Ramanujan

Ramanujan

@DHMO lol again stucked

DHMO

@Ramanujan wrong in the second line

Ramanujan

What?

DHMO

13:44

@Ramanujan wrong in the second line

Ramanujan

Ergh what's wrong there?

Sorry, I got it,big mistake from me

Regretting

Will be solving agoin

@DHMO now what?

DHMO

@Ramanujan expand

Ramanujan

@DHMO now?????????

DHMO

14:00

@Ramanujan can u show the steps above?

Ramanujan

DHMO

$a^2(c-b)$ is wrong

@Ramanujan

Ramanujan

From which step?😥

DHMO

from the step containing $a^2(c-b)$

oh, never mind

$b^2(a-b)$ is wrong

from the step containing $b^2(a-b)$

Ramanujan

OK , so I should take it $b^2(a-c)$

I will come in next five - 10 minutes after solving whole(I think SE is keep distracting me) iam usually better

Ramanujan

14:21

DHMO

@Ramanujan group the remaining four terms into two groups

$b^2a-b^2c+c^2b-c^2a$

wait

Make it $b^2a-c^2a+c^2b-b^2c$

then group the first two and the last two

brb

Ramanujan

@DHMO can't see any further,trying to stand on your shoulders for next steps

@DHMO are you there?

@DHMO are you there??

Secret

14:47

Geometric finiteness

In geometry, a group of isometries of hyperbolic space is called geometrically finite if it has a well-behaved fundamental domain. A hyperbolic manifold is called geometrically finite if it can be described in terms of geometrically finite groups. == Geometrically finite polyhedra == A convex polyhedron C in hyperbolic space is called geometrically finite if its closure C in the conformal compactification of hyperbolic space has the following property: For each point x in C, there is a neighborhood U such that all faces of C meeting U also pass through x (Ratcliffe 1994, 12.4). For example, every...

Please describe exactly what is "well behaved"

SteamyRoot

I would assume this is in the article...?

Ramanujan

@DHMO U there?

DHMO

15:04

@Ramanujan start from here lol

$a^2(c-b)+b^2a-b^2c+c^2b-c^2a = a^2(c-b)+b^2a-c^2a+c^2b-b^2c = a^2(c-b)+(b^2-c^2)a+cb(c-b)$

here

Ramanujan

15:16

@DHMO did you got that right?

DHMO

@Ramanujan maybe

Ramanujan

Ah,i also need to go through another topics

YOUSEFY

15:26

Hello

Does any one here specialize in Graph Theory

I would like to know about inequlity in graph theory that says the minimum degree of subgraph is greater than the ratio of the subgraph (assuming that a graph has at least one edge)

I want to know how to prove it?

Balarka Sen

@Secret Not sure about the point of asking a question that's answered in the article you linked...

YOUSEFY

Ok

give me a second

Lozansky

I think I have a singularity in my tooth

Secret

They don't really say what is meant by "well -behaved for a fundamental domain, they only give an example of polyhedra in hyperbolic space

Balarka Sen

@Secret Please read carefully.

Secret

15:40

> A discrete group G of isometries of hyperbolic space is called geometrically finite if it has a fundamental domain C that is convex, geometrically finite, and exact

> A convex polyhedron C in hyperbolic space is called geometrically finite if its closure C in the conformal compactification of hyperbolic space has the following property:
For each point x in C, there is a neighborhood U such that all faces of C meeting U also pass through x

> A hyperbolic manifold is called geometrically finite if it has a finite number of components, each of which is the quotient of hyperbolic space by a geometrically finite discrete group of isometries

Unless the convex polyhedron case somehow is referring to the fundemental domain...?

all other examples keep referencing back to the term "geoemtrically finite"

Balarka Sen

Fundmental domain of a discrete group of isometries of the hyperbolic space is always ever a polyhedron.

Secret

ok, didn't knew that...

I thought it might be something more fancy or abstract

DHMO

abc.net.au/news/2016-10-19/…

Evinda

16:13

Hello @AndrewThompson
Are you familiar with harmonic functions?

Andrew Thompson

@Evinda I know the definition.

Evinda

That is my question:

SteamyRoot

Sometimes it is hard not to cry when correcting homework

Evinda

Let $u(x,y)$ be a harmonic function in $\mathbb{R}^2$. I want to compute $u(0,0)$ given that $u|_{x^2+y^2=1}=\sin{\phi}+1$, where $x= \cos{\phi}, y=\sin{\phi}, \phi \in [0,2 \pi)$.

From a known theorem we have that the only solution of our problem is the following:

$ u(x,y)=\frac{1-(x^2+y^2)}{2 \pi} \int_{x^2+y^2=1} \frac{(\xi_2+1)}{|x-\xi|^2} dS =\frac{1-(x^2+y^2)}{2 \pi} \int_{x^2+y^2=1} \frac{(\xi_2+1)}{(x- \xi_1)^2+ (y-\xi_2)^2} dS $

Is it right so far? If so, can we compute the integral?

Andrew Thompson

Sorry, no idea.

Evinda

16:24

A ok

Hey @TedShifrin
Do you make have an idea?

That is the theorem that I mean:

$(\star) \Delta u=0 \text{ in } B_R(0), u|_{\partial{B_R(0)}}=\phi $

($ B_R(0)$ is a sphere with radius $R$)

Theorem: Let $ \phi \in C^0(\partial{B_R}(0))$ , then the function $ u(x)=\frac{R^2-|x|^2}{w_n R} \int_{\partial{B_R}} \frac{\phi(\xi)}{|x-\xi|^n} dS $ is the only (classical) solution of $(\star)$.

($ w_n $ is the area of the unit ball in $ \mathbb{R}^n $)

Andrew Thompson

Well, you've written $x = \cos \phi$, $y = \sin \phi$. Then $x^2 + y^2 = 1$, right?

Evinda

Yes @AndrewThompson

Andrew Thompson

Then your integral is zero, no?

You write $$\frac{1 - (x^2 + y^2)}{2\pi}$$ before the integral. That's 0.

Evinda

Yes? I have also thought of this, but I thought that we could not use the x and y at this point because we integrate over $x^2+y^2=1$. So we integrate over the whole space, right?

Andrew Thompson

I'm assuming you mean to say that you integrate over $x^2 + y^2 = 1$, hence the polar coordinates at that point. If that's the case I still have no idea. If it is as you've written, i.e. x = cos theta and y = sin theta, then it is 0.

Evinda

16:28

@AndrewThompson So have I applied the theorem correctly?Or have I done something wrong at this part: $\int_{\partial{B_R}} \frac{\phi(\xi)}{|x-\xi|^n} dS $ ?

SuperMan

Hello

Can you give me an advice how to find the solutions of x?
https://www.wolframalpha.com/input/?i=x%5E2*2%5E(x%2B1)%2B2%5E(%7Cx-3%7C%2B2)%3Dx%5E2*2%5E(%7Cx-3%7C%2B4)%2B2%5E(x-1)

This works:
https://goo.gl/F0yL0R

Evinda

16:44

@Apeiron Hi... are you greek?

Lozansky

16:56

How is the difficulty of an introductory course in probability theory compared to say a basic course in complex analysis or vector calc?

s.harp

17:16

is it just me or do most gravatars suggest an anti-clockwise spiral?

Mike Miller

Mine's clockwise.

s.harp

my mind sees it as anti-clockwise^

Mike Miller

meh

s.harp

the parts pointing in a clockwise direction
make it look like they are trailing behind
$==== \qquad <--$
$\qquad || \quad <--$

Ajax Edm

17:47

Hey guys you have time to help me a bit out in linear algebra?

Alessandro

Just state your question instead of asking if you can ask!

Ajax Edm

Well this is the question i have . I dont really fully understand the answer and need to see how to do thus

thi math.stackexchange.com/questions/1975603/…

I mean i do not grasp what i have to show

Alessandro

You have to describe those 3 rings as simply as possible, showing that they are isomorphic to well known rings as it's done in the answer is probably the easiest approach

Ajax Edm

18:04

@Alessandro I mean i just do not fully understand how i mean how would i do this for the first example a)

Alessandro

Do you agree that your groups are $\mathbb{Z}/n\mathbb{Z}$ for $n=2,3,4$ and that they are cyclic so their automorphisms are decided by the image of a generator?

Ajax Edm

hmm i think that's precisely what i do not understand .. I mean I only understand what a ring / group is... and the different types of morphism...

Alessandro

ok, so you don't know what a cyclic group is?

Ajax Edm

nope ... I had only a brief introduction to groups rings and morphisms ... and this is part of my linear algebra course

Alessandro

hm, ok, you should notice that in any of your $3$ groups all of the elements of the group can be obtained by "adding" $f$ to itself repeatedly (where "adding" refers to the group operation described by your table)

Sylent Nyte

18:14

hey guys, need some help with understanding matrices, i have this matrix.. $\begin{pmatrix} 6 & -3 \\ -4 & 2 \end{pmatrix}$, I know its determinent is $0$ but i dont understand how the plane is mapped onto the line $2x+3y=0$

Alessandro

In the $4$th, for example, $f=f$, $g=f+f$, $h=g+f=f+f+f$ and $e=h+f=f+f+f+f$

Suppose now that you have an automorphism $\phi$, if you know $\phi (f)$ can you determine what's, say, $\phi (g)$ using the fact I stated above and the fact that $\phi$ is an automorphism?

arctic tern

@SylentNyte solve for A and B in the equation A(6x-3y)+B(-4x+2)=0

Sylent Nyte

@arctictern thanks, I played around a little and figured it out for myself a little while ago, but thank you either way :)

Sylent Nyte

18:45

using matrices, if I want to rotate(R) and then enlarge(E), is it RE or ER? and why?

arctic tern

When you apply a matrix A to a vector x, it's Ax right?

So then if AB denotes matrix multiplication, applying AB to x yields ABx, or in other words A(Bx). That means applying B to x first (giving Bx) and then applying A (that is, applying A to Bx giving A(Bx)).

Sylent Nyte

19:09

what if its three..?

would it be A(B(Cx))?

dsillman2000

20:27

Hi guys

Lozansky

@TedShifrin You there?

dsillman2000

I think he might be AFK

Jeff

Question? From the left of a vertical asymptote, the function goes to infinity. From the right it intersects the VA with an open circle. Is this an infinite or jump discontinuity? (My answer: infinite (or maybe "both" (no, I think infinite again))).

arctic tern

"jump discontinuity" means the left and right limits exist (as real numbers) and aren't equal, right?

Jeff

@arctictern yes (well, don't have the def. handy). therefore it's not jump.

arctic tern

20:36

> don't have the def handy

well, step 1 is always to identify the problem

Jeff

(true that)

Lozansky

Let $f$ be a Möbius transformation such that $f(1) = \infty$ and $f$ maps the imaginary axis onto the unit circle $|w|=1$. What is the value of $f(-1)$?

arctic tern

@Lozansky any guesses?

Lozansky

So I think that the transformation is given by $w = f(z) = \frac{z+1}{z-1}$ which gives $f(-1) = 0$

arctic tern

well, there's more than one such transformation

but your guess is correct

Lozansky

20:40

@arctictern Okay, I also wish to see this by symmetry

@arctictern Are you familiar with the symmetry principle?

arctic tern

probably

maybe we could draw the real axis, imaginary axis, and unit circle, then use the fact mobius transformations send generalized circles to generalized circles and preserve angles

Lozansky

Let $C_z$ be a line or circle in the $z$-plane, and let $w=f(z)$ by any Möbius transformation. Then two points $z_1$ and $z_2$ are symmetric with respect to $C_z$ if and only if their images $w_1 = f(z_1), w_2 =f(z_2)$ are symmetric with respect to the image of $C_z$ under $f$

And since $1$ and $-1$ are symmetric with respect to the imaginary axis...

Their images should too?

arctic tern

if $f$ sends the imaginary axis to the unit circle, then it must send the real axis and unit circle to two perpendicular lines that intersect the unit circle perpendicularly. it preserves the intersection points, so if f(1)=infinity that forces f(-1)=0

Lozansky

@arctictern Does it send the unit circle back to the imaginary axis?

arctic tern

let me get some paper so I can draw this

Lozansky

20:54

Cool, thanks

arctic tern

no, not necessarily

Lozansky

Why not? It's one-to-one

arctic tern

doesn't really matter. it can rotate axes around the origin.

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$Mathematics$ Mathematics