Results 1 to 7 of 7

Math Help - Complex Locus

  1. #1
    Member
    Joined
    Jul 2009
    Posts
    132

    Complex Locus

    z = x + iy, which satisfies

    z^2 + z^ -2 = 8

    Find the locus of P in terms of x and y.

    Is the general aproach to get a equantion in x and y?

    If so, i tried this and ive got z^4 - 8z^2 + 1 = 0
    Do i just sub in x + iy as z and expand to get locous?

    If not what do i do?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by Lukybear View Post
    z = x + iy, which satisfies

    z^2 + z^ -2 = 8

    Find the locus of P in terms of x and y.

    Is the general aproach to get a equantion in x and y?

    If so, i tried this and ive got z^4 - 8z^2 + 1 = 0
    Do i just sub in x + iy as z and expand to get locous?
    You could do that, but it's likely to be rather messy.
    If not what do i do?
    I would suggest substituting w := z^2 first. If you do that you'll find that w_{1,2}=4\pm \sqrt{15} are the only solutions. Now solve the two equations (x+iy)^2=4\pm \sqrt{15} and you are done.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,814
    Thanks
    703
    Hello, Lukybear!

    z \,=\, x + iy\text{, which satisfies: }\;z^2 + z^{-2} \:=\: 8

    \text{Find the locus of }P\text{ in terms of }x\text{ and }y.

    The problem is not clearly stated.
    I assume that we want the locus of z.

    We have: . z^4 - 8z^2 + 1 \:=\:0

    Quadratic Formula: . z^2 \:=\:\dfrac{8\pm\sqrt{60}}{2} \:=\:\dfrac{8\pm2\sqrt{15}}{2} \:=\:4 \pm\sqrt{15} .[1]


    We note that: . (\sqrt{5}\pm \sqrt{3})^2 \;=\;5 \pm 2\sqrt{15} + 3 \;=\;8 \pm 2\sqrt{15} \;=\;2(4 \pm \sqrt{15})

    . . Hence: . 4 \pm \sqrt{15} \:=\:\dfrac{(\sqrt{5} \pm \sqrt{3})^2}{2}


    Then [1] becomes: . z^2 \;=\;\dfrac{(\sqrt{5}+\sqrt{3})^2}{2}

    . . Hence: . z \;=\;\pm\left(\dfrac{\sqrt{5}\pm\sqrt{3}}{\sqrt{2}  }\right) . . . all real numbers.


    The locus is four vertical lines:

    . . \begin{array}{ccccc}<br />
x &=& \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}} & \approx & 2.806\\ \\[-3mm]<br />
x &=& \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{2}} & \approx & 0.356\\ \\[-3mm]<br />
x &=& \dfrac{-\sqrt{5} + \sqrt{3}}{\sqrt{2}} & \approx & \text{-}0.356\\ \\[-3mm]<br />
x &=& \dfrac{-\sqrt{5} - \sqrt{3}}{\sqrt{2}} & \approx & \text{-}2.806 \end{array}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2009
    Posts
    132
    Ok thanks. Much easier than i had anticpiated.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by Soroban View Post
    The locus is four vertical lines:

    . . \begin{array}{ccccc}<br />
x &=& \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}} & \approx & 2.806\\ \\[-3mm]<br />
x &=& \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{2}} & \approx & 0.356\\ \\[-3mm]<br />
x &=& \dfrac{-\sqrt{5} + \sqrt{3}}{\sqrt{2}} & \approx & \text{-}0.356\\ \\[-3mm]<br />
x &=& \dfrac{-\sqrt{5} - \sqrt{3}}{\sqrt{2}} & \approx & \text{-}2.806 \end{array}
    No, I don't believe that that's quite right. Going back to z^2=4\pm\sqrt{15} and setting z := x+iy, we get (x+iy)^2=x^2+2ixy-y^2=4\pm\sqrt{15}.
    Comparing real- and imaginary parts of the left and right hand side we see that we must have xy=0 and x^2-y^2=4\pm\sqrt{15}.

    But x=0 is not possible, because then the second of these two equations would not have a (real) solution y.

    Setting y=0 to satisfy the first equation gives the solutions x_{1,2,3,4}=\pm\sqrt{4\pm\sqrt{15}} of the second equation.

    Hence I get z_{1,2,3,4}=\pm\sqrt{4\pm\sqrt{15}} four isolated points (and not four vertical lines).
    Last edited by Failure; August 30th 2010 at 07:09 AM. Reason: fixed wrong signs in expansion of (x+iy)^2=
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jul 2009
    Posts
    132
    So what is the locus?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by Lukybear View Post
    So what is the locus?
    As I wrote, it is just the four points z_{1,2,3,4}=\pm\sqrt{4\pm \sqrt{15}} on the real axis. This agrees with the solution Soroban gave you, with the small difference that y cannot be arbitrary, but must be =0.

    If you take z:= \sqrt{4+\sqrt{15}}+i\cdot 2, for example, which would lie on one of those vertical axes Soroban found, you get that the left hand side of the original equation evaluates to z^2+1/z^2\approx 3.90+i\cdot 11.14, which is rather too far away from the value of the right hand side, 8.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex numbers, how to describe and sketch a locus?
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: May 7th 2011, 05:50 PM
  2. complex number locus
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 28th 2008, 02:10 AM
  3. complex no. + locus again
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 28th 2008, 05:19 AM
  4. locus of complex no. question no. 2
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 22nd 2008, 11:07 AM
  5. locus of complex no.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 19th 2008, 11:06 PM

Search Tags


/mathhelpforum @mathhelpforum