complex argument

[thorny badger]

How to solve this one?

[mint wadi]

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[thorny badger]

By conjugate multiplication

x^2-y^2+2ixy/(x^2+y^2)

(x^2-y^2)/(x^2+y^2)+ 2ixy/x^2+y^2

arctan|(2xy)/(x^2-y^2)|

[prime nest]

,rotate

[prime nest]

$\frac{x+iy}{x-iy}=\frac{x^2-y^2+2ixy}{x^2+y^2}=\frac{x^2-y^2}{x^2+y^2}+i\frac{2xy}{x^2+y^2}$\
\
$z=a+bi\implies \arg z=\tan^{-1}\qty(\frac{b}{a})$.

@thorny badger

[muted yoke]

Romans 12:19

[thorny badger]

Acha

Us hisab se arc tan (2xy/x^2-y^2)

[thorny badger]

Itna toh ho gya tha bhai upr Maine bhi main ye angle wali problem aa rhi h mujhe@prime nest

[prime nest]

abe answer aa gya hai mera

☠️

oh nahi ayaa

L

😭

[thorny badger]

Socha thoda balak

Answer ni aaya h vo galat hain A

@sweet bolt

[prime nest]

ruk ja do min

[thorny badger]

[prime nest]

@sweet bolt pleas confirm (b)

[thorny badger]

Yes B

Vhi toh seekna h munhe 😆

Exam me toh aa laga aata

a

[sweet bolt]

hi

[thorny badger]

Can you explain angle of arctan?

Ye negative value aayegi because of |y|>|x|

matlab arctan(-)

[sweet bolt]

yes

[thorny badger]

Ab aage?

[sweet bolt]

Matlab?

[prime nest]

usne $\arctan\qty(\frac{2xy}{x^2-y^2})$

[muted yoke]

Romans 12:19

[prime nest]

ko simplify krna h taaki option jaisa kuch aaye

[thorny badger]

,w value of arctan(-1/√3)

[prime nest]

aag lgadi

[muted yoke]

Results provided by WolframAlpha

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Requested by arjunn5589

[thorny badger]

,w arz of (-1-√3i)

[muted yoke]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by arjunn5589

[sweet bolt]

fakesolve it lol

[thorny badger]

Via putting value

-1+√2i

[wooden seal]

do you also mix english and hindi in written language? 😄

[sweet bolt]

nop

[thorny badger]

Papers are provided bilingual they know that i am going to ask questions on discord with foreigners

[thorny badger]

Bro are you in India?

Do we not have papers in Hindi/English/Bengali?

[sweet bolt]

he didn't ask about papers

[thorny badger]

Ohh

Fer unko space problem hogi??

[stray mica]

Note that Arg(z1/z2) = arg(z1) - arg(z2) and arg(conj(z)) = -arg(z).

[thorny badger]

Her problem@stray mica is with angle

[stray mica]

Well, the argument is the angle. Or did you mean something else?

[thorny badger]

I guess you didn't see the thing which i wrote and where i am stuck with the main problem with selecting the current option

[thorny badger]

@stray mica

[stray mica]

Yes?

[thorny badger]

I am sorry for pinging you

But i guess it is just language barrier

I am stuck on finding how to select the quadrant and the angle

[stray mica]

Let's see. We have:
Arg(z/conj(z)) = arg(z) - arg(conj(z)) = 2arg(z)
Now, we note that |y/x| > 1, so z is in this region.

Thus, z^2 is in this region (not purely real and negative, though).

Let's look at how we determine the argument.

I don't think there's a unique answer. It depends on the sign of y.