Greatest Moduli Complex Number Solution of Equation

[crimpedupcan]

I would very much appreciate any help with this problem.

Homework Statement

Find the greatest value of the moduli of the complex numbers z satisfying the equation

|z - \frac{4}{z}| = 2

The Attempt at a Solution

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

\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} = 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 z.

 

[crimpedupcan]

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:
\left|z\right| - \left|\frac{4}{z}\right| ≤ \left|z - \frac{4}{z}\right|
∴ \left|z\right| - \left|\frac{4}{z}\right| ≤ 2
i.e. \left|z\right|^{2} - 2\left|z\right| - 4 ≤ 0
i.e. \left(\left|z\right| - 1\right)^{2} ≤ 5
i.e. \left|z\right| ≤ \sqrt{5} + 1

So the largest value of the modulus of z is \sqrt{5} + 1

 

[Ray Vickson]

I would very much appreciate any help with this problem.

Homework Statement

Find the greatest value of the moduli of the complex numbers z satisfying the equation

|z - \frac{4}{z}| = 2

The Attempt at a Solution

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

\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} = 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 z.

You would have to maximize a^2 + b^2, 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:
z = r \cos(\theta) + i r \sin(\theta), \; \frac{4}{z} = \frac{4}{r}\cos(\theta) - i \frac{4}{r} \sin(\theta), so
|z - 4/z|^2 = \left(r - \frac{4}{r}\right)^2 \cos^2(\theta) + <br /> \left( r + \frac{4}{r}\right)^2 \sin^2(\theta) = 4.
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