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Complex number problem, could the problem be wrong?

  1. Aug 24, 2011 #1
    Hi!

    I was presented with this problem:

    Let z +[itex]\neq -1[/itex] be a complex number with |z|=1.
    Show that [itex]\frac{z-1}{z+1}[/itex] is an imaginary number.
    I am getting nowhere with the algebra, but i did notiec one thing:

    let z=1+0b and the equation says [itex]\frac{z-1}{z+1}[/itex] = [itex]\frac{0}{2}[/itex] = 0, which is not an imaginary number..

    Is the problem incorrect?

    Thanks!
     
  2. jcsd
  3. Aug 24, 2011 #2

    Mark44

    Staff: Mentor

    What you have is a valid counterexample to the statement.
     
  4. Aug 24, 2011 #3
    I think you have the question wrong. Let me try and make it more interesting:

    If [itex]|z|=1[/itex], then show [itex]\frac{z-1}{z+1}[/itex] is pure imaginary except when [itex]z=\pm 1[/itex]
     
  5. Aug 24, 2011 #4
    Thanks for the replies.

    The problem itself seems pretty straightforward on second look.

    Let z = a + ib.
    [itex]\frac{(a-1) +ib}{(a+1) +ib}[/itex]

    We multyply by the complex conjugate and get

    [itex]\frac{((a-1)+ib)((a+1)-ib)}{(a+1)^2+b^2}[/itex],

    [itex]\frac{(a-1)(a+1) -ib(a-1) +ib(a+1) +b^2}{(a+1)^2 +b^2}[/itex]

    And since |z| = [itex]\sqrt{a^2 +b^2}[/itex]=1 [itex]\Rightarrow[/itex] a^2 + b^2 = 1
    , the a^2 + b^2 -1 casel and all that is left is

    [itex]\frac{i*2b}{(a+1)^2+b^2}[/itex],
     
  6. Aug 24, 2011 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi mariush! :smile:

    this a + ib stuff is very undignified :redface:

    try treating complex numbers as numbers, not number-pairs

    rewrite |z| = 1 as [itex]z\bar{z} = 1[/itex]

    then what is the condition Re(f(z)) = 0 as an equation in z and [itex]\bar{z}[/itex] ? :wink:
     
  7. Aug 24, 2011 #6
    Tim :)

    I agree it looks horrible. I have found it helpful to think of the number-pairs as vectors (hencec thinking in number pairs) and since it's my first day whit these complex numbers I haven't gotten to the section that discusses de Moivres formula ++ yet.

    I don't think i fully understand what you ment by the last. Maybe you could explain?
     
  8. Aug 24, 2011 #7

    tiny-tim

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    Science Advisor
    Homework Helper

    hi mariush! :smile:

    (no, i'm not suggesting you use de moivre)

    if a function f(z) is purely imaginary (ie Re(f(z)) = 0), then what is the relation between f(z) and [itex]f(\bar{z})[/itex] ? :wink:
     
  9. Aug 24, 2011 #8
    I don't think so. 0 is a pure imaginary number. Its real part is zero, right?
     
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