1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    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


    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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
     
  4. Sep 13, 2011 #3

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Write [itex]z=a+bi[/itex].

    Can you calculate

    [tex]\frac{z-1}{z+1}[/tex]

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

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

    [tex]\frac{a-1+bi}{a+1+bi}=\frac{(a-1+bi)(a+1-bi)}{(a+1+bi)(a+1-bi)}[/tex]

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    OK, so it is

    [tex]\frac{2b}{2a+1}i[/tex]

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Well, what is the argument of

    [itex]0+ci[/itex]

    ???
     
  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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    c is just a real number.

    What angle does i make with the x-axis?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Complex Analysis - Value of Arg[(z-1)/(z+1)] between -pi and pi
Loading...