Results 1 to 15 of 15
Like Tree2Thanks
  • 2 Post By Gusbob

Math Help - Complex numbers

  1. #1
    Newbie
    Joined
    Jun 2013
    From
    Australia
    Posts
    9

    Complex numbers

    Hi guys. I need some help with I guess a relatively simple question. Could someone please show me with workings the following ?
    If the modulus of z+(1/z)=2. What is z?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    4,166
    Thanks
    762

    Re: Complex numbers

    Hey Yeah01.

    Hint: Can you calculate an expression for the modulus in terms of x and y (for x + iy) or in terms of r and theta (polar co-ordinates)?

    (In other words, if you have z + 1/z, what is the modulus in terms of x and y if z = x + iy)?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2013
    From
    Australia
    Posts
    9

    Re: Complex numbers

    Hi chiro,
    Hope your having a good night thanks for your response. I've tried that still must be doing something wrong.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    4,166
    Thanks
    762

    Re: Complex numbers

    Hint: the modulus of z is SQRT(x^2 + y^2). If you are dealing with 1/z then the modulus is 1/SQRT(x^2 + y^2) since if z = r*e(i*theta) then 1/z = 1/r * e^(-i*theta) and 1/r is the modulus of 1/z.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2013
    From
    Australia
    Posts
    9

    Re: Complex numbers

    Thanks chiro. Let me give it a go
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jun 2013
    From
    Australia
    Posts
    9

    Re: Complex numbers

    Still no luck. Don't know what I'm doing wrong
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jun 2013
    From
    Australia
    Posts
    9

    Re: Complex numbers

    Do u think u can help me out chiro?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jun 2013
    From
    Australia
    Posts
    9

    Re: Complex numbers

    Hi I got x = 1/4-y and y=1/4-x. What now? Sm I on the right track?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Jun 2013
    From
    Australia
    Posts
    9

    Re: Complex numbers

    Hi I got x = 1/4-y and y=1/4-x. What now? Sm I on the right track?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,965
    Thanks
    1785
    Awards
    1

    Re: Complex numbers

    Quote Originally Posted by Yeah01 View Post
    Hi I got x = 1/4-y and y=1/4-x. What now? Sm I on the right track?
    I don't think at you are.

    As you can see here, this problem has a rather nice answer.

    But when I tried it, the algebra is really messy. From that link, I tried those solutions and they work.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Jun 2013
    From
    Australia
    Posts
    9

    Re: Complex numbers

    Thanks. What did yr working look like. The answer is good but really want work out how to do it.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,965
    Thanks
    1785
    Awards
    1

    Re: Complex numbers

    Quote Originally Posted by Yeah01 View Post
    Thanks. What did yr working look like. The answer is good but really want work out how to do it.
    Well I really have absolutely no idea how to work it out algebraically.
    I taught complex variables on and off for over thirty years. But I have never seen a more difficult problem in this category of problems than this one is.

    Clearly z=\pm 1 are solutions simply by inspection.

    But how one gets the other eight solution remains a puzzle to me.

    There may well be a very clever trick that I have not thought of.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member
    Joined
    Jan 2008
    Posts
    588
    Thanks
    87

    Re: Complex numbers

    I believe the most intuitive way to do this is to look at the pre image of the Joukovski's transformation z\mapsto z+ \frac{1}{z}. This is a conformal map, and there are some nice ways you can deal with this. I also believe it can be cranked out algebraically in an acceptably nice way. I'm working on that now.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    Jun 2013
    From
    Australia
    Posts
    9

    Re: Complex numbers

    Wow. Thanks. Looking forward to seeing if someone can do the math.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Super Member
    Joined
    Jan 2008
    Posts
    588
    Thanks
    87

    Re: Complex numbers

    Let w=\frac{1}{2}\left(z+\frac{1}{z}\right). Then the condition in the question is equivalent to saying |w|=1. If w=\pm 1, you'll recover the trivial answers z=\pm 1

    Exercise: If |w|=1 and w\not= \pm 1, show that \arg\left(\frac{w-1}{w+1}\right) is \frac{\pi}{2} or \frac{3\pi}{2}. That is, \frac{w-1}{w+1} is purely imaginary.

    HINT: Writing w=x+iy, you should find that \frac{w-1}{w+1}=\frac{2iy}{(x+1)^2+y^2}

    Fun facts: w-1=\frac{1}{2z}(z-1)^2 and w+1=\frac{1}{2z}(z+1)^2

    Hence \frac{w-1}{w+1}=\left(\frac{z-1}{z+1}\right)^2 =ai for some real number a, with some restrictions on a.

    Now you need some bounds on |a|= \left|\frac{w-1}{w+1}\right|=\left|\frac{2iy}{(x+1)^2+y^2}\right  | since you had the restriction |w|=\sqrt{x^2+y^2}=1, y\not=0

    I didn't have time to finish calculating the bounds, but note that |a|=\left|\frac{\sqrt{2}-1}{\sqrt{2}+1}\right| and |a|= \left|\frac{\sqrt{2}+1}{\sqrt{2}-1}\right| gives the easier solutions in Plato's wolfram alpha link. It comes out by solving \left(\frac{z-1}{z+1}\right)^2 =ai.

    Good luck with that. I'll work on the bounds later if I have time.
    Last edited by Gusbob; June 1st 2013 at 09:24 PM.
    Thanks from Yeah01 and Plato
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. raising complex numbers to complex exponents?
    Posted in the Advanced Math Topics Forum
    Replies: 10
    Last Post: March 25th 2011, 11:02 PM
  2. Replies: 1
    Last Post: September 27th 2010, 04:14 PM
  3. Replies: 2
    Last Post: February 7th 2009, 07:12 PM
  4. Replies: 1
    Last Post: May 24th 2007, 04:49 AM
  5. Complex Numbers- Imaginary numbers
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 24th 2007, 01:34 AM

Search Tags


/mathhelpforum @mathhelpforum