how to prove

[lone torrent]

Amp(z)+amp(z')=2kπ where k belong to (-π,π]

[viscid needle]

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[round crest]

by z' do you mean the conjugate of z

[lone torrent]

Yes

[round crest]

then $\text{amp }z+\text{amp }\bar{z}=0$

[half warren]

waffle

[round crest]

because

If $z=a+ib$ then $\text{amp }z=\tan^{-1}\qty(\frac ba)$ and\$\text{amp }\bar{z}=\tan^{-1}\qty(-\frac ba)=-\tan^{-1}\qty(\frac ba)$

[half warren]

waffle

[lone torrent]

Are you sure?

[round crest]

hmm yeah

[lone torrent]

https://youtu.be/Ri8BormIBTo?si=azy9sMXAuYZCbaEC

[round crest]

isn't amplitude defined as the principal argument

[half warren]

waffle

[sterile gate]

what does amp mean

[lone torrent]

The angle which a complex number made with x axis

[sterile gate]

so thats the argument...l

so trying to prove $arg(z)+arg(\bar{z})=2k{\pi}, where k\in(-\pi,\pi]$

[half warren]

;(

[sterile gate]

let $z=a+bi$

[half warren]

;(

[sterile gate]

thus $\bar{z}=a-bi$

[half warren]

;(

[sterile gate]

now $arg(z)=arctan(b/a)$ and $arg(\bar{z})=arctan(-b/a)$

[half warren]

;(

[lone torrent]

Yes

Right as someone wrote

tan^-1(b/a)+tan^(-1)(-b/a)

[sterile gate]

same thing

[lone torrent]

,w arctan(b/a)+arctan(-b/a)

[half warren]

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Requested by arjunn5589

[sterile gate]

yeah

but we have to consider that $k\in(-\pi,\pi]$

[half warren]

;(

[cinder peak]

You also didnt quote the result properly

[sterile gate]

okay this is interesting...

[cinder peak]

from the video, it claims it equals $2\pi k$ for $k\in\mathbb{Z}$

[half warren]

Omegabet_

[lone torrent]

....-2π,0,2π,4π....

[cinder peak]

yes, that's $2\pi\mathbb{Z}$

[half warren]

Omegabet_

[cinder peak]

but yeah, it's 0 cause t+(-t)=0

the angle 0 is just coterminal to everything in 2piZ

[sterile gate]

yes thats what i was going to da

say*

[cinder peak]

so looks like they're taking arg not principal, and just taking all the elements from the equivalence class

[cinder peak]

you were echoing what was already said

[sterile gate]

okay

[cinder peak]

$z=re^{i\theta}\to\text{arg}(z)=\theta+2\pi\mathbb{Z}$

[half warren]

Omegabet_

[cinder peak]

so $\arg(z)+\arg(\overline{z})=(\theta+2\pi\mathbb{Z})+(-\theta+2\pi\mathbb{Z})=2\pi\mathbb{Z}$ as sets

[half warren]

Omegabet_

[sterile gate]

thats the conclusion ig

[lone torrent]

I see

[sterile gate]

its just using coterminal angles is all

[lone torrent]

@cinder peak

@sterile gate

[cinder peak]

yeah

and?

[sterile gate]

what

you want me to prove it??

[cinder peak]

I know you love posting pictures without words, but learn to be coherent

[lone torrent]

i am asking is it true?

[sterile gate]

...

[cinder peak]

duh?

[sterile gate]

prove it yourself first...

[lone torrent]

here in the picture

[sterile gate]

we are not answer machines

we are here to help

if you want to check validity just prove it

yourself

[lone torrent]

amp(z)+amp(z')=amp(1/z)

So amp(1/z)=2kπ?

[sterile gate]

uh

z=a+bi

try to get the imaginary part on the numerator

[cinder peak]

clearly not always

[sterile gate]

$\frac{1}{z}=\frac{1}{a+bi}$

[half warren]

;(

[sterile gate]

try to prove it by conjugating the fraction

[lone torrent]

It will be (a-bi)/(a^2+b^2)

[sterile gate]

yes

what do you know a^2+b^2 is

[lone torrent]

It is |z|^2

And positive always

Slept?@sterile gate

[lone torrent]

I am asking if both identities you giys are saying is true then why I cannot say they are equal to each other

@cinder peak

[cinder peak]

please.... learn to read

[lone torrent]

I see

[cinder peak]

where does this say arg(z*) = -arg(z)+arg(1/z)?

[lone torrent]

In the question they fixed its domain

So it will be 0?

[cinder peak]

there's no domain to discuss, it's not a function

$\arg(\overline{z})=-\arg(z)$ is trivial to prove

[half warren]

Omegabet_

[cinder peak]

$z=re^{i\theta}\to\overline{z}=re^{-i\theta}=re^{i(-\theta)}$

[half warren]

Omegabet_

[cinder peak]

QED

$z=re^{i\theta}\to\frac{1}{z}=\frac{1}{re^{i\theta}}=\frac{1}{r}e^{-i\theta}$

[half warren]

Omegabet_

[cinder peak]

QED

[sterile gate]

lol

[cinder peak]

$\arg(\overline{z})$ is both $-\arg(z)$ and $\arg(\frac{1}{z})$

[half warren]

Omegabet_

[cinder peak]

you've written nonsense

[lone torrent]

Huh

[cinder peak]

$\overline{z}-z$ isnt $a-bi$

[half warren]

Omegabet_

[lone torrent]

I wrote z'=-z

[cinder peak]

it's $(a-bi)-(a+bi)=-2bi$

[half warren]

Omegabet_

[cinder peak]

no you didnt

[lone torrent]

I meant z' only

[cinder peak]

then write that

[lone torrent]

Sorry remove -z

-z=-a-bi

[cinder peak]

yes

there's no -z so idk why you care about -z

[sterile gate]

no

what

[cinder peak]

Like please for the love of god stop doing math until you have learned to communicate. You consistently struggle to communicate with us

[sterile gate]

not true?

[cinder peak]

it's true

distributive property still holds in C.

[sterile gate]

-z doesnt equal a-bi...

-z = -a-bi...

[cinder peak]

you too, learn to read

they wrote -a-bi

[sterile gate]

for z?

i cant read that

[lone torrent]

Lol

[lone torrent]

How is it equal? 1/r is not same as r

Or we care only for angle not for amplitude r

Thanks i got it now. I was right yes we care only angle

[cinder peak]

we're taking argument.

so yes, modulus is irrelevant

[lone torrent]

Right

So in the OP they wrote the angle lies k belongs to (-π,π] and you have wrote 2kπ so it Is clearly 0

.....-4π,-2π,0,2π,4π...

[cinder peak]

they didnt write k in (-pi,pi]

from the video you yourself watched and posted in the thread

(nowhere do they even have a k so...)

[lone torrent]

No no

[cinder peak]

................

[cinder peak]

do I need to spell out how 2pi*k isnt k, or can you see that for yourself?

but also $2\pi\mathbb{Z}\cap(-\pi,\pi]={0}$, so refer back to the very start where everyone said it was $0$.

[half warren]

Omegabet_

[cinder peak]

(so they're taking amp to be principal arg, but same difference)

[lone torrent]

Yes

[cinder peak]

yes

[lone torrent]

.....-4π,-2π,0,2π,4π...

[cinder peak]

oh you're back to babbling, good to know

[lone torrent]

Omega

Vitamin

[cinder peak]

do you still have questions or just going to be a fool?

[lone torrent]

No question

[cinder peak]

+close