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Greatest Moduli Complex Number Solution of Equation

  1. Feb 25, 2012 #1
    I would very much appreciate any help with this problem.

    1. The problem statement, all variables and given/known data

    Find the greatest value of the moduli of the complex numbers [itex]z[/itex] satisfying the equation

    [itex]|[/itex][itex]z[/itex] - [itex]\frac{4}{z}[/itex][itex]|[/itex] = 2

    3. The attempt at a solution

    I tried letting [itex]z[/itex] = [itex]a+bi[/itex] and going from there, but I ended up with this really large equation:

    [itex]\left(\frac{a\left(a^{2} + b^{2} - 4\right)}{a^{2} + b^{2}}\right)^{2} + \left(\frac{b\left(a^{2} + b^{2} + 4\right)}{a^{2} + b^{2}}\right)^{2}[/itex] = 4

    and I don't know how to simplify it. And even if I did simplify it, I don't know how I would find the greatest value of the moduli of [itex]z[/itex].
     
  2. jcsd
  3. Feb 29, 2012 #2
    Okay so nobody commented (maybe I posted it in the wrong place) but I have since found a solution and thought I would post it, in case anyone got curious:

    Using the triangle inequality:
    [itex]\left|z\right| - \left|\frac{4}{z}\right| ≤ \left|z - \frac{4}{z}\right|[/itex]
    ∴ [itex]\left|z\right| - \left|\frac{4}{z}\right| ≤ 2[/itex]
    i.e. [itex]\left|z\right|^{2} - 2\left|z\right| - 4 ≤ 0[/itex]
    i.e. [itex]\left(\left|z\right| - 1\right)^{2} ≤ 5[/itex]
    i.e. [itex]\left|z\right| ≤ \sqrt{5} + 1[/itex]

    So the largest value of the modulus of [itex]z[/itex] is [itex]\sqrt{5} + 1[/itex]
     
  4. Feb 29, 2012 #3

    Ray Vickson

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    You would have to maximize [itex]a^2 + b^2[/itex], subject to a, b being restricted by your equation above. That problem involves calculus (at the level of Calculus II). A somewhat easier way is to look at z and 1/z in the (x,y) plane, using polar coordinates:
    [tex] z = r \cos(\theta) + i r \sin(\theta), \; \frac{4}{z} = \frac{4}{r}\cos(\theta) - i \frac{4}{r} \sin(\theta),[/tex] so
    [tex] |z - 4/z|^2 = \left(r - \frac{4}{r}\right)^2 \cos^2(\theta) +
    \left( r + \frac{4}{r}\right)^2 \sin^2(\theta) = 4. [/tex]
    We want the largest r for which there will be a θ that solves that equation. By setting up the Lagrange multiplier problem we find that either θ = 0 or θ = π/2. When we try θ = π/2 we find there is no real value of r that works. Therefore, we must have θ = 0; that is, z must lie on the positive x-axis. Now the problem becomes one of solving the equation x - 4/x = 2, from which we find x = sqrt(5) + 1, as you gave.

    Your solution was OK, but it overlooked one important consideration: you need to know that it is possible to find z in which your triangle inequality becomes an equality. Of course, that is true if we take z = x + i*0 with x = sqrt(5)+1, because in that case z - 4/z = x - 4/x = 2 exactly.

    RGV
     
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