Results 1 to 11 of 11

Math Help - Simplify complex equation

  1. #1
    Senior Member
    Joined
    Apr 2010
    Posts
    487

    Simplify complex equation

    The question:
    |\sqrt{w^2 - 4} + w| = 2 such that w is a complex number

    I tried using the identity |z|^2 = z\overline{z} to get a solution, but it appears to create a mess. :/

    Any assistance would be great.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,656
    Thanks
    1480
    \displaystyle |\sqrt{w^2 - 4} + w| = 2

    \displaystyle |\sqrt{w^2 - 4} + w|^2 = 2^2

    \displaystyle (\sqrt{w^2 - 4} + w)^2 = 4.

    Now try to solve for \displaystyle w...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Prove It View Post
    \displaystyle |\sqrt{w^2 - 4} + w| = 2

    \displaystyle |\sqrt{w^2 - 4} + w|^2 = 2^2

    \displaystyle (\sqrt{w^2 - 4} + w)^2 = 4.

    Now try to solve for \displaystyle w...
    I'm not sure this will work since |a|^2 = a \overline{a}, not a^2, when a is complex.

    @OP: I'd be inclined to substitute w = x + iy and take it from there, messy as it's likely to be .... (Where has the question come from?)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Apr 2010
    Posts
    487
    Quote Originally Posted by mr fantastic View Post
    I'm not sure this will work since |a|^2 = a \overline{a}, not a^2, when a is complex.

    @OP: I'd be inclined to substitute w = x + iy and take it from there, messy as it's likely to be .... (Where has the question come from?)
    It's part of a complex analysis question:

    Show that the function f defined by f(z) = z^2 + z^{-2} maps the unit circle onto the interval [-2, 2]
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by Glitch View Post
    It's part of a complex analysis question:

    Show that the function f defined by f(z) = z^2 + z^{-2} maps the unit circle onto the interval [-2, 2]
    If z is on the unit circle then |z|^2 = z \overline{z} = 1, so z^{-1} = \overline{z}.

    Then f(z) = z^2 + \overline{z}^2 = (z + \overline{z})^2 - 2 z \overline{z} = (z + \overline{z})^2 -2

    So...
    Last edited by awkward; March 19th 2011 at 12:46 PM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Apr 2010
    Posts
    487
    Quote Originally Posted by awkward View Post
    If z is on the unit circle then |z|^2 = z \overline{z} = 1, so z^{-1} = \overline{z}.
    Sorry, I'm not sure how you came to the conclusion that z^{-1} = \overline{z}.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,656
    Thanks
    1480
    Because on the unit circle, \displaystyle |z|^2 = 1 \implies z\overline{z} = 1 \implies \overline{z} = \frac{1}{z}.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Apr 2010
    Posts
    487
    Oh, right. >_<

    I need stronger coffee.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Apr 2010
    Posts
    487
    Quote Originally Posted by awkward View Post
    (z + \overline{z})^2 -2
    OK, so I've worked out what you've done so far. This is what I've done:

    Let z = x + iy

    (x + iy + x - iy)^2 - 2
    (2x)^2 - 2
    4x^2 - 2

    So unless I'm mistaken, this is a parabola. I'm fairly sure this isn't the correct solution. :/
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by Glitch View Post
    OK, so I've worked out what you've done so far. This is what I've done:

    Let z = x + iy

    (x + iy + x - iy)^2 - 2
    (2x)^2 - 2
    4x^2 - 2

    So unless I'm mistaken, this is a parabola. I'm fairly sure this isn't the correct solution. :/
    Since (x,y) is on the unit circle, -1 \leq x \leq 1.

    So then what is the range of 4x^2 - 2?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member
    Joined
    Apr 2010
    Posts
    487
    Ahh, thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simplify Complex Fraction
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 12th 2011, 12:36 PM
  2. Simplify this complex fraction?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 30th 2010, 08:40 PM
  3. complex numbers simplify
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 18th 2009, 06:21 AM
  4. simplify a complex fraction
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 10th 2008, 03:55 AM
  5. Simplify the complex fraction
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 25th 2008, 10:39 AM

Search Tags


/mathhelpforum @mathhelpforum