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user223679

04:50

chat.stackexchange.com/users/46626/waffles-crazy-peanut

I invited him here. @Mann

Mann

06:14

Added him fb nvm

^^

@user223679

user223679

06:30

Facebook? @Mann

Mann

Yes.

user223679

I guess he declined my invitation. :/

Mann

:OOO

Hahahaha :D

I didn't got replied to my messages either :|

0

Q: complex nos in ellipse.

I was practising some ques on ellipses when I came a criss this question: If normal at four points $(x_1,y_1)$..... on the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ are concurrent then find the value of $$(x_1+x_2+x_3+x_4)\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\right)...

complex-numbers

Dinesh

Yes

Sorry connection prob

Mann

Ah np , so did you let $z=re^{i\theta}$ ?

In polar form?

Dinesh

06:34

No. r=1

Mann

Why would r=1 :O

Well there would be two problem with it, it wouldn't really make an ellipse in an ellipse r is a function of theta

Check this

Ellipse

In mathematics, an ellipse is a curve on a plane surrounding two focal points such that the sum of the distances to the two focal points is constant for every point on the curve. As such, it is a generalization of a circle, which is a special type of an ellipse that has both focal points at the same location. The shape of an ellipse (how 'elongated' it is) is represented by its eccentricity, which for an ellipse can be any number from 0 (the limiting case of a circle) to arbitrarily close to but less than 1. Ellipses are the closed type of conic section: a plane curve that results from the ...

Dinesh

In the original eqn we have sin x. and cosx. . So I let z=cos+i sin

This z i s not in polar corrdinate

Mann

But $z$ specify what in your equation?

Dinesh

We have cos x and sin x in equation right

Mann

Yes

I haven't made it or tested it however

Dinesh

06:37

So. I put cos + i sin =some z

So I replaced cos by (z +1/z)/2 and so

Mann

Hmm can you send me the equation you have?

Dinesh

OK

Mann

Hi robert!

user223679

Btw, if you got the 4th degree eqn, the 1st bracket is sum of the roots while the second is (summation of three at a time)/(product of roots) @Dinesh

Mann

I believe he has already done it like that.

Dinesh

06:40

U r wrong

We want $x_1$

Which is cos theta

Which is (z+1/z)/2

So 1/x_1 means z^2/(z^2+1)

Mann

Isnt your equation like $-(a^2-b^2)x'^4+2ha^2(a^2-b^2)x'^3+x'^2(....)-2a^4h(a^2-b^2)x'+a^6h^2$

user223679

Oh! You got the eqn in z and not x. I didn't read the question properly.

Dinesh

Yeah

In z

Mann

Can x' has value 3?

think

Dinesh

And in eq z^2 term does not occur

Mann

06:43

Or can x' attain values greater than 1

Dinesh

How are you even commenting on the range of x

Mann

Well you defined z= cos + i sin doesn't that represent a unit circle?

So if x attains any value greater than 1 it's over.

or even under

Dinesh

Well don't put z graphically

Mann

Uh ok let's suppose x is just that nvm

How did you simplify?

after

Dinesh

I didn't put z=cos+i sin. Instead I put cos =(z+1/z)/2 and so on. Notice the difference between the two claims

I got the equation as

Mann

06:47

Let me get link for mathjax sec

math.ucla.edu/~robjohn/math/mathjax.html

Dinesh

Z^4(a^2-b^2)+z^3(2iyb-2ax)+z(2ax+2iyb)-(a^2-b^2)

Mann

Open it and render mathjax

Start*

$\frac{1}{2}$

Works

Dinesh

$$Z^4(a^2-b^2)+z^3(2iyb-2ax)+z(2ax+2iyb)-(a^2-b^2$$

Mann

Yes works does it for you?

Dinesh

Nope but I will manage. Hope u see me equation

Mann

06:52

Yes, Pretty much. just drag the link of start mathjax into your bookmark

Come on this chat and click the bookmark

Done

user223679

3

Q: 2,4-dinitrophenyl hydrazine test

I found this reaction over here: which illustrates the carbonyl test but looks fuzzy. Assuming that the R in the product is a typo for H, I couldn't conserve the number of Nitrogen atoms. What is the correct reaction? Is this correct? Is there a possibility of a methyl substitution ...

organic-chemistry reaction-mechanism analytical-chemistry erratum

@Mann

Mann

Cool

Let me see

Dinesh

I am on mobile so can't do it. But I'll type in latex . Its fine

Mann

Exactly what i was thinking xD

Ow mobile

I see

I don't really see the origin of this statement :'(, can not understand where you got it from. What was your initial question "If the normals at four points $(x_i,y_i)$ on the ellipse are concurrent then value of $x_1+x_2+x_3+x_4$ $(1/x_1 +1/x_2$ etc?

@user223679 , the methyl was probably biggest typo

xD

Dinesh

Yeah that was it

Mann

07:00

So why would you want to do it in complex number? You would still have to extract the "real part" of it in the solution leading to same equation.

user223679

@Mann The reason why I gave up.

Mann

Hahha D: Askiitan are sometimes weird

Giving so many things without any reason or proof

GG

user223679

@Mann I gave up on that Complex number question, and regarding askiitians, I gave up on it last year ;)

Mann

Ahahahahhahaha I see. xD

Ofcourse, weird things on it i see.

Still trying the complex one though

I still believe x should be represent by $\frac{z+\bar z}{2}$

user223679

@Mann Can you create chat rooms ?

Mann

07:04

Yes.

user223679

With restricted entry?

Mann

But it would create too much mess this is fine.

Nope I guess D:

Never tried

Dinesh

@Mann u are right . x is $\frac{z+\bar z}{2}$ except for the fact that its $\frac{z+\frac{1}{z}}{2}$

Mann

Did you use equation of normal in parametric form?

Wait

user223679

The reason for this is that, this room has been converted into a discussion room while it was meant to be a continuation of the comments of the original answer to ADG's question. @Mann

Mann

07:08

I know haha, but not a problem. It just works fine.

user223679

I don't mind it as long as someone does not come in and yells at us for making this room a mess ;)

Mann

Ahahaha, but there are already too many messages above anyway it wouldn't matter.

Or we can always change name

user223679

chat.stackexchange.com/rooms/info/23542/…

Mann

OH damn , okay new room

Indeed

user223679

This room has no owners. It belongs to that answer and the name most probably can't be changed.

I suggest we call in a mod. He can shift the useless messages to our new room.

Mann

07:11

Uh not really the best idea, not all messages are so nice in here are they? x

Oh, I see @Dinesh now what had you mean, tried this in parametric form

user223679

But they wont be relevant over here atleast. Better shift them over there.

Mann

Uh i guess you are right.

user223679

You all continue with the complex number question, while I go in search of a mod with lots of free time ;)

Dinesh

Yeah. Now you got

Mann

So i have my equation all in terms of cos

4th power

But still it's perfect in here why would you still have to use complex number instead of cos

Dinesh

07:16

Bcz u can't proceed easily with cos equation. So instead convert everything in just z and we get a 4th power eq in z

Mann

It's very easy in actual

Multiply the roots of cos with a and you got x_1 x_2 x_3 x_4

Dinesh

Ohh I got it. I did it this way because I solved every conormal ques in z and got easy results. Thanks anyways

Mann

Weird, I never tried that though! Must try these somedays haha

Dinesh

Yeah u easily get the results this way. Like if parametric angles are a,b,c,d then a+b+c+d=odd multiple of pi.... and so on

Mann

:OOO, Interesting. I think i sometimes keep things to mainstream aa.

Wait a+b+c+d wasn't that the condition for con-cyclic points?

Dinesh

07:26

Nope. A,b, c,d are angled

Angles

Mann

Yes angles of con cyclic points on ellipse sorry.

Dinesh

Moreover we can use complex nos easily to show if 4 points are concyclic

Mann

:O

Dinesh

Any four general points

Wait. I'll tell ya

Mann

I don't remember much about complex number it seems, :| Gotta do much revision. Was it with rotation theorem?

Dinesh

07:29

z_1 be a complex no. Then we can rotate z as $z_2=z_1 cis(\theta)$

Pureferret

Hello?

Mann

Oh yes, I remember that rotation theorem np.

HI!

Pureferret

Someone rang?

user223679

Me!

Pureferret

Ah ha!

user223679

07:29

You got a minute? @Pureferret

Pureferret

Not sure I'd I can up and lift chats elsewhere

But I have a minute

Mann

Cool sorted then :D

user223679

So you can shift these messages? @Pureferret

Mann

I believe the opposite points must be diameters of an circle! @Dinesh right?

or not

Dinesh

Not necessary

Mann

07:31

Nevermind.

Nope

checked it

Dinesh

Do u know how to do it easily by complex NOs ( how to show 4 points are con cyclic)

Mann

Nope, even If i would, by now i would have forgotten that. :(

Pureferret

@user223679 which ones and where?

Dinesh

If a, b, c, d are 4 points and a,b,c,d are come led nis then if they are con cyclic then

$\frac{(a-b)(c-d)}{(a-c)(b-d)}$ is purely real

user223679

@Mann Did you create a new room?

Mann

07:36

Umm no, wait let me do it

Pureferret

It can be done by a room owner I think...

user223679

@Pureferret The problem is, these messages are owned by the answer/comments from where it started. So no owners.

Mann

$Mathematics$ We love math, physics, chemistry and …

Random Talk. If you don't understand the talk, hardluck!

Nice one @Dinesh !

user223679

@Pureferret I know its too much but, all messages after this one:

Pureferret

Whosoever owns this room, click on the room▼ link then select the comments

user223679

07:38

13 hours ago, by user223679

Hello guys!

Mann

I'd like to prove that :D

Pureferret

@user223679 no, I said room owners. The comments from the answer comments will be owned by the person who commented

Who owns the room?

Heck let me make that you @user223679

I'm in mobile so I can't ctrl-click them

Mann

Yes! @user223679 for president! D:

Ow

user223679

@Mann Check whether you are the owner -_-

Mann

How do i do that. **

Pureferret

07:41

Click info

Mann

Seems not, I have never owned anything in my life D:

Nope

Doesn't show any owner

user223679

Yes! Thats why I thought of calling in a mod. Maybe they can shift/overide these stuff.

Mann

@Dinesh , can't get a nice proof. do you know D:

Pureferret

@StackExchange Boom. By a landslide victory @user223679 is now president

@user223679 if you click on room▼ and then in move/delete you should see instructions

Mann

Cool. I actually owned something, interesting. Thanks
!

Pureferret

07:44

Tbh, you could just rename this chat

user223679

@Pureferret Thanks. But I just realized that a person on mobile can't multiselect(you had the same problem). So @Mann please shift them.

Pureferret

And with that, I shall disappear into the night

Mann

Hail pureferret

user223679

@MannJust changed it.

Mann

← 301 messages moved from We Love Maths

Done

Dinesh

07:47

@Mann use this: if 4 points are con cyclic then angles in same arc ate equal

And then use concept of complex slope

See if u can do from that

user223679

@Dinesh The other room please.

Mann

OH

This in indeed better

Nope, I was never so good with complex number d:

user223679

@Mann You made the new room, I just changed everything about this room!! Let me see if I can restore stuff back.

Mann

Got it!

user223679

@Mann @Dinesh Go to the new room please.

Mann

07:51

Or maybe not

ok

Left other room.

>Still trying to prove concyclicity
>Failing
>Fml

user223679

@Mann There are few messages left to shift, the ones of Dinesh. Shift them here. I tidied up the other room in case someone goes there, and make me the owner over here too ;)

Mann

Ok, but i already left other room **

How to make owner

user223679

$Mathematics$ Discussion between ADG and Mann

From the comments of math.stackexchange.com/questions/1240775/…

Firstly, go back over there and shift the 5-6 messages.

Mann

← 13 messages moved from Discussion between ADG and Mann‌

Ok

user223679

Great!! Now go to info and [add user] as owner.

@Mann ?

Mann

08:01

Trying **

Not able to woot

Where to do it

user223679

Go over here(chat.stackexchange.com/rooms/info/23760/we-love-math) , scroll down. Under owners, on the RHS there will be [add user]. Search my name.

Mann

@Dinesh , be mine savior and bring me forth to thine light. **

user223679

@Mann Hello!! Did you try?

Mann

Trying lol wait screenshot

user223679

Okay. Go to [access] tab. You are currently on [general].

Go over [here](http://chat.stackexchange.com/rooms/info/23760/we-love-math?tab=access) and user223679
user223679 scroll down. Under owners, on the RHS there will be [add user]. Search my name.

Mann

08:08

Did i made it!?

Wait screen

user223679

Ah

I need 100rep to be a owner. Else a mod has to make me, like Purefeet did. You be the owner till that time. I will soon get rep and then you make me.

Mann

Ah yes, just answer some nice questions np

Here easy one

0

Q: Finding the points of intersection on a circle

Before addressing my issues, below is the question from a past examination paper along with a diagram I dre in order to facilitate readers. 3(a) A circle has center $C(5, 8)$ and radius $\sqrt{29}$ (ii) The circle cuts the y-axis at $R$ and $Q$, where $Q$ is above $R$. Find the coordin...

trigonometry circle

user223679

@Mann Yes. Gtg. see you later.

Mann

Uh cya, I should study or maybe eat then x

user223679

@Mann I will gain rep soon. But right now, I have to go. Cya.

Mann

09:54

math.stackexchange.com/questions/1281627/… xD

@user223679