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.

  • #31
OK, now the \frac{2b}{2a+1} is just a real number, so we're dealing with a multiple of i.

So our value could be 3i, 4i, 6i or some other multiple of i. What is the argument of a multiple of i?
 
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  • #32
micromass said:
OK, now the \frac{2b}{2a+1} is just a real number, so we're dealing with a multiple of i.

So our value could be 3i, 4i, 6i or some other multiple of i. What is the argument of a multiple of i?

pi/2, right?
 
  • #33
cooljosh2k2 said:
pi/2, right?

Not really, because we can also have negative multiples. -i and -3i are also multiples.
 
  • #34
micromass said:
Not really, because we can also have negative multiples. -i and -3i are also multiples.

well, pi/2 for positive multiples of i or I am > 0, or -pi/2 for negative multiples of i, or I am < 0
 
  • #35
Indeed! That's it!
 
  • #36
micromass said:
Indeed! That's it!

WOW! Thank you so much for helping and putting up with me!
 

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