The maximum value of |z|, when z satisfy the condition |z + 3/z|=3 ?

Question

This question is from complex numbers.

Answer 1

Vaibhav and Richard,

 

The solution given by Richard above shows a correct approach to the problem, but makes one small but crucial misstep.  In the line where the modulus of the expression (z + 3/z) is computed, the given formula is not quite correct.  Letting |z|=R, and computing instead the square of the modulus, it should read:

 

|z + 3/z|² = |Re^iθ + 3/Re^iθ|²

               = (R + 3/R)²cos²(θ) + (R - 3/R)²sin²(θ) 

               = R² + (6cos(2θ)) + 9/R²

 

and upon equating this with 3² = 9, we obtain

 

R⁴ + (6cos(2θ) - 9)R² + 9 = 0.

 

In the last equation, which is also quadratic in R², note that the maximal solution will occur when the coefficient of the R²-term is as negative as possible.  This occurs at the minimal value of cos(2θ), which occurs at θ = π/2, and which is of course -1.  The resulting equation is then

 

R⁴ -15R² + 9 = 0,

 

whose maximal solution is 

 

R² = (15 + √189)/2

 

and hence the maximal value for R = |z| is 

 

√( (15 + √189)/2 ).

 

By the way, we now also know a complex number z that takes this maximal modulus solution, namely z = i√( (15 + √189)/2 ).

 

Regards,

Hassan

Answer 2

Since z is complex and we are interested in |z|,  represent z as   |z| exp(iΘ).     Then

 

   | z + 3/z | =   sqrt( |z|² + 9/ |z|²).      The condition then reads

 

     sqrt( |z|² + 9/ |z|²) = 3.    this is an equation for a the real quantity |z|² .

 

    Squaring both sides and rearranging yields a quadratic equation for |z|²

    

   ( |z|² )² - 9 |z|²  +9 = 0.       The quadratic formula can be used to find the the two solutions to this equation.

  The larger one is     |z|²  = ( 9 + sqrt(45) ) /2.      So

 

    |z|  =  sqrt (  (9 + sqrt(45)) /2 ).