1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Maximum Value Complex Function

  1. Jun 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the maximum value of |(z-1)(z+1/2)| for |z|≤1.

    2. Relevant equations
    Calculus min/max concepts?

    3. The attempt at a solution
    Let f(z)=|(z-1)(z+1/2)|. Observe f(z) is the product of 2 analytic functions on |z|≤1, g(z)=z-1 and h(z)=z+1/2. Therefore f(z) is analytic on |z|≤1. Since f(z) is analytic on |z|≤1, it is continuous on the same domain.

    f'(z)=2z-1/2.

    Critical points at z=1 and 1/4.

    f(1)=0
    f(1/4)=9/16

    The maximum value of |(z-1)(z+1/2)| for |z|≤1 is 9/16.

    Is this close? I suspect I'm missing something fundamental as this doesn't use any complex analysis content.
     
    Last edited: Jun 29, 2012
  2. jcsd
  3. Jun 29, 2012 #2
    is the question asking for the modulus of (z-1)(z+1/2)?
     
  4. Jun 29, 2012 #3
    because in that case, it wont follow like a normal equation. modulus(z) is not a complex differentiable function. since all the components of your functions only affect your real part, try using i, or -i
     
  5. Jun 29, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The complex analysis concept you are missing is the maximum modulus theorem. Parametrize the boundary and look for the maximum there.
     
  6. Jul 1, 2012 #5
    How does this look?

    Let |z|≤1 be a domain, D. Let f(z)=(z-1)(z+1/2). Observe f(z) is the product of 2 analytic functions: g(z)=z-1 and h(z)=z+1/2. Therefore f(z) is analytic on D. Since f(z) is analytic on D, it is also continuous on D. By Maximum Modulus Theorem, the max|f(z)| occurs on the boundary of D.

    f(z)=(z-1)(z+1/2). Let z=e. f(e)=(e-1)(e+1/2)=e2iθ-e/2-1/2=cos(2θ)+isin(2θ)-cos(θ)/2-isin(θ)/2-1/2.

    Then |f(e)|=√(cos(2θ)-cos(θ)/2-1/2)2+(sin(2θ)-sin(θ)/2)2. d/dθ|f(e)|=0→sin(θ)(8cos(θ)+1)=0→θ=k1π for k1=0,±1,... OR θ=cos-1(-1/8)+2k2π for k2=0,±1,...

    |f(ei*0)|=0
    |f(ei*cos-1(-1/8))|=√25/64+(sin(cos-1(-1/8))-sin(cos-1(-1/8))/2)2>0, and is the maximum value.
     
    Last edited: Jul 1, 2012
  7. Jul 1, 2012 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Looks pretty ok to me. I'm getting a max at theta=arccos(-1/8) as well.
     
  8. Jul 2, 2012 #7
    Thanks for the help, Dick. I really appreciate it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Maximum Value Complex Function
Loading...