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

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SUMMARY

The discussion focuses on calculating the argument of the complex expression Arg[(z-1)/(z+1)], where z is a point on the unit circle (|z|=1). It is established that Arg[(z-1)/(z+1)] equals π/2 when the imaginary part of z (Im(z)) is greater than 0, and -π/2 when Im(z) is less than 0. The participants derive this by expressing (z-1)/(z+1) in terms of its real and imaginary components and confirming that it is purely imaginary, leading to the conclusion about the argument's values.

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  • Understanding of complex numbers and their representation (a + bi).
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  • Knowledge of the argument function Arg(z) and its geometric interpretation.
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  • Learn about the geometric interpretation of the argument of a complex number.
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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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