Maximum Value Complex Function

1. Jun 29, 2012

ChemEng1

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. Jun 29, 2012

tt2348

is the question asking for the modulus of (z-1)(z+1/2)?

3. Jun 29, 2012

tt2348

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

4. Jun 29, 2012

Dick

The complex analysis concept you are missing is the maximum modulus theorem. Parametrize the boundary and look for the maximum there.

5. Jul 1, 2012

ChemEng1

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
6. Jul 1, 2012

Dick

Looks pretty ok to me. I'm getting a max at theta=arccos(-1/8) as well.

7. Jul 2, 2012

ChemEng1

Thanks for the help, Dick. I really appreciate it.