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

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Let Arg(w) denote that value of the argument between -π and π (inclusive). Show

    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

    3. 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? Im stuck here. Please help
  2. jcsd
  3. Sep 13, 2011 #2


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    Homework Helper

    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.

  4. Sep 13, 2011 #3
    Write [itex]z=a+bi[/itex].

    Can you calculate


    in terms of a and b?? In particular, can you make the denominator real?
  5. Sep 13, 2011 #4
    Is this what im 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 im wrong, what am i doing wrong?
  6. Sep 13, 2011 #5
    No, that's not what I get. Just calculate


    What do you get??
  7. Sep 13, 2011 #6
    Ok, i get: (a2+b2+2bi-1)/(a2+b2+2a+1)

    Is that right?
  8. Sep 13, 2011 #7
    Yes, and what is [itex]a^2+b^2[/itex]??? (hint: z is on the unit circle).

    Can you show that [itex]\frac{z-1}{z+1}[/itex] is purely imaginary?? What are the consequences for the arg?
  9. Sep 13, 2011 #8
    Wow, am i rusty. I am so embarrassed to say that i dont know. I may be wrong, but i think i faintly remember [itex]a^2+b^2[/itex]= 1. But i dont think its right. As for your other questions, i dont know. Im sorry, im an idiot.
  10. Sep 13, 2011 #9
    Yes, [itex]a^2+b^2=1[/itex]. So what does that mean for your equation?
  11. Sep 13, 2011 #10
    Wow, im shocked i remembered that lol. It would equal 2bi/(2a +1).

    Thanks for helping me through it btw.
  12. Sep 13, 2011 #11
    OK, so it is


    Now calculate the argument. With the arctangent if you prefer.
  13. Sep 13, 2011 #12
    How do i do that with 2b and 2a+1?
  14. Sep 13, 2011 #13
    Well, what is the argument of


  15. Sep 13, 2011 #14
    0? My textbook barely even touches on the argument, thats why im so lost.
  16. Sep 13, 2011 #15
    Are you still there? Im so confused.
  17. Sep 13, 2011 #16
    What does your book say that the definition of the Arg is??
  18. Sep 13, 2011 #17
    "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 = Im z/|z| ."
  19. Sep 13, 2011 #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?
  20. Sep 13, 2011 #19
    I dont know what "c" is, am i supposed to come up with a number? Does ci = sinθ?
  21. Sep 13, 2011 #20
    c is just a real number.

    What angle does i make with the x-axis?
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