Maximum Value Complex Function

[ChemEng1]

Homework Statement

Find the maximum value of |(z-1)(z+1/2)| for |z|≤1.

Homework Equations

Calculus min/max concepts?

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.

 

[tt2348]

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

 

[tt2348]

because in that case, it won't 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

 

[Dick]

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

 

[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^iθ. f(e^iθ)=(e^iθ-1)(e^iθ+1/2)=e^2iθ-e^iθ/2-1/2=cos(2θ)+isin(2θ)-cos(θ)/2-isin(θ)/2-1/2.

Then |f(e^iθ)|=√(cos(2θ)-cos(θ)/2-1/2)²+(sin(2θ)-sin(θ)/2)². d/dθ|f(e^iθ)|=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(e^i*0)|=0
|f(e^{i*cos-1(-1/8)})|=√25/64+(sin(cos^-1(-1/8))-sin(cos^-1(-1/8))/2)²>0, and is the maximum value.

 

[Dick]

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^iθ. f(e^iθ)=(e^iθ-1)(e^iθ+1/2)=e^2iθ-e^iθ/2-1/2=cos(2θ)+isin(2θ)-cos(θ)/2-isin(θ)/2-1/2.

Then |f(e^iθ)|=√(cos(2θ)-cos(θ)/2-1/2)²+(sin(2θ)-sin(θ)/2)². d/dθ|f(e^iθ)|=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(e^i*0)|=0
|f(e^{i*cos-1(-1/8)})|=√25/64+(sin(cos^-1(-1/8))-sin(cos^-1(-1/8))/2)²>0, and is the maximum value.

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

 

[ChemEng1]

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