Complex Analysis - Value of Arg[(z-1)/(z+1)] between -pi and pi

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Homework Help Overview

The discussion revolves around the argument of a complex function, specifically Arg[(z-1)/(z+1)], where z is constrained to the unit circle (|z|=1). Participants explore the implications of this constraint on the values of the argument, particularly focusing on the conditions under which the imaginary part of z is positive or negative.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the argument using the properties of complex numbers, including the use of real and imaginary parts. There are attempts to express the function in terms of a and b, the real and imaginary components of z, and to simplify the expression to identify the argument.

Discussion Status

The conversation has progressed through various calculations and clarifications, with participants questioning their understanding of the argument and its geometric interpretation. Some have provided guidance on how to approach the problem, while others express uncertainty about their calculations and concepts.

Contextual Notes

There is an emphasis on understanding the definition of the argument in relation to the unit circle, and participants are encouraged to reflect on their knowledge of trigonometric relationships and the properties of complex numbers.

cooljosh2k2
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Homework Statement


Let Arg(w) denote that value of the argument between -π and π (inclusive). Show
that:

Arg[(z-1)/(z+1)] = { π/2, if Im(z) > 0 or -π/2 ,if Im(z) < 0.

where z is a point on the unit circle ∣z∣= 1


The Attempt at a Solution



First, i know that Arg(w) = arctan(b/a) and that Arg(z/w) = Arg(z) - Arg(w).

So Arg[(z-1)/(z+1)] = Arg(z-1) - Arg(z+1).

But then what do i do? I am stuck here. Please help
 
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Note that z is a point of the unit circle. What are its real and imaginary parts? You can find the real and imaginary parts of (z-1)/(z+1) if you multiply both the numerator and denominator with the conjugate of the denominator: z*+1.

ehild
 
Write z=a+bi.

Can you calculate

\frac{z-1}{z+1}

in terms of a and b?? In particular, can you make the denominator real?
 
micromass said:
Write z=a+bi.

Can you calculate

\frac{z-1}{z+1}

in terms of a and b?? In particular, can you make the denominator real?

Is this what I am supposed to do:

(z-1)/(z+1) = (a + bi -1)/(a+bi +1) = (a2 - b2 +2abi -1)/ (a+bi+1)2

Is that right? If so, what do i do next, or if I am wrong, what am i doing wrong?
 
cooljosh2k2 said:
Is this what I am supposed to do:

(z-1)/(z+1) = (a + bi -1)/(a+bi +1) = (a2 - b2 +2abi -1)/ (a+bi+1)2

Is that right? If so, what do i do next, or if I am wrong, what am i doing wrong?

No, that's not what I get. Just calculate

\frac{a-1+bi}{a+1+bi}=\frac{(a-1+bi)(a+1-bi)}{(a+1+bi)(a+1-bi)}

What do you get??
 
micromass said:
No, that's not what I get. Just calculate

\frac{a-1+bi}{a+1+bi}=\frac{(a-1+bi)(a+1-bi)}{(a+1+bi)(a+1-bi)}

What do you get??

Ok, i get: (a2+b2+2bi-1)/(a2+b2+2a+1)

Is that right?
 
Yes, and what is a^2+b^2? (hint: z is on the unit circle).

Can you show that \frac{z-1}{z+1} is purely imaginary?? What are the consequences for the arg?
 
micromass said:
Yes, and what is a^2+b^2? (hint: z is on the unit circle).

Can you show that \frac{z-1}{z+1} is purely imaginary?? What are the consequences for the arg?

Wow, am i rusty. I am so embarrassed to say that i don't know. I may be wrong, but i think i faintly remember a^2+b^2= 1. But i don't think its right. As for your other questions, i don't know. I am sorry, I am an idiot.
 
cooljosh2k2 said:
Wow, am i rusty. I am so embarrassed to say that i don't know. I may be wrong, but i think i faintly remember a^2+b^2= 1. But i don't think its right. As for your other questions, i don't know. I am sorry, I am an idiot.

Yes, a^2+b^2=1. So what does that mean for your equation?
 
  • #10
micromass said:
Yes, a^2+b^2=1. So what does that mean for your equation?

Wow, I am shocked i remembered that lol. It would equal 2bi/(2a +1).

Thanks for helping me through it btw.
 
  • #11
cooljosh2k2 said:
Wow, I am shocked i remembered that lol. It would equal 2bi/(2a +1).

Thanks for helping me through it btw.

OK, so it is

\frac{2b}{2a+1}i

Now calculate the argument. With the arctangent if you prefer.
 
  • #12
micromass said:
OK, so it is

\frac{2b}{2a+1}i

Now calculate the argument. With the arctangent if you prefer.

How do i do that with 2b and 2a+1?
 
  • #13
Well, what is the argument of

0+ci

?
 
  • #14
micromass said:
Well, what is the argument of

0+ci

?

0? My textbook barely even touches on the argument, that's why I am so lost.
 
  • #15
Are you still there? I am so confused.
 
  • #16
What does your book say that the definition of the Arg is??
 
  • #17
micromass said:
What does your book say that the definition of the Arg is??

"Arg z, the argument of z, defined for z ≠ 0, is the angle which the vector (originating
from 0) to z makes with the positive x-axis. Thus Arg z is defined (modulo
2π) as that number θ for which
cos θ = Re z/|z|
; sin θ = I am z/|z| ."
 
  • #18
OK, so the arg is the angle that the vector makes with the x-axis. Now, what angle does ci make with the x-axis?
 
  • #19
micromass said:
OK, so the arg is the angle that the vector makes with the x-axis. Now, what angle does ci make with the x-axis?

I don't know what "c" is, am i supposed to come up with a number? Does ci = sinθ?
 
  • #20
c is just a real number.

What angle does i make with the x-axis?
 
  • #21
micromass said:
c is just a real number.

What angle does i make with the x-axis?

I don't know. You make it sound so simple, but I am so confused. I am sorry.

arcsin(Im(z)/|z|)?
 
  • #22
Pick a piece of Paper. Draw an x-axis and a y-axis on it. Find the point represented by i. Connect i to 0. What is the angle between the line segment that you've drawn and the x-axis?? Just look at it.
 
  • #23
micromass said:
Pick a piece of Paper. Draw an x-axis and a y-axis on it. Find the point represented by i. Connect i to 0. What is the angle between the line segment that you've drawn and the x-axis?? Just look at it.

Isnt i on the y-axis? which would make the angle 0.
 
  • #24
What is the angle between the x-axis and the y-axis?
 
  • #25
micromass said:
What is the angle between the x-axis and the y-axis?

OMG, i forgot you said in relation to the x axis. 90 degrees. *hits head*
 
  • #26
cooljosh2k2 said:
OMG, i forgot you said in relation to the x axis. 90 degrees. *hits head*

or pi/2
 
  • #27
Yes!

The argument of i is just the angle between i and the x-axis. So, what is the argument of i??
 
  • #28
micromass said:
Yes!

The argument of i is just the angle between i and the x-axis. So, what is the argument of i??

pi/2!
 
  • #29
And what is the argument of -i??
 
  • #30
micromass said:
And what is the argument of -i??

-pi/2! but how do i go from (2bi)/(2a+1)? do i just ignore the (2b/2a+1)?
 

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