Find Max/Min of |f(z)| at z=2: Maximum Modulus Theorem & Minimum Modulus Theorem

[wany]

Homework Statement

Suppose that $$f(z)=(z-1)(z-4)^2.$$ Find the lines through z=2 on which |f(z)| has a relative maximu, and those on which |f(z)| has a relative minimum, at z=2.

Homework Equations

Maximum Modulus Theorem And Minimum Modulus Theorem

The Attempt at a Solution

So I know that f(z) is not differentiable at z=1. So I could take the domains (-5,1) and (5,1). Since we know that the maximum and minimum must be located on the boundary of the domain.
However, I am not sure how to do this or if this is right.

 

[HallsofIvy]

You seem to be completely confused here. f certainly is differentiable at z= 1. For any z, the derivative is f'(z)= (z- 4)^2+ 2(z- 1)(z- 4) which is 25 at z= -1.

Further, I have no idea what you could mean by "lines through z=2 on which |f(z)| has a relative maximum". When z= 2, |f(2)|= f(2)= 4, a constant.

 

[wany]

Well that was the word for word problem from the book, and the book also said that |f| is not differentiable at z=1 and has a saddle point at z=2.

 

[HallsofIvy]

Yes, |f(z)| is NOT differentiable at z= 1 but that is not what you said. You said that f(z) was not differentiable at z= 1 and it is. Also, perhaps not stated in the problem but said elsewhere, I feel sure, is that this is a problem in complex variables. "The line z= 2" is the line where z= 2+ ix for any real number x. Then $f(z)= (z- 1)(z- 4)^2$$= (2- ix-1)(2- ix- 4)^2$$= (1- ix)(-2- ix)^2$. Write out |f(z)| as a function of x, find the derivative, set it equal to 0. etc.

 

[wany]

Alright, sorry my bad I didn't explain correctly. First of all, yes this is a complex problem. So we are assuming that $$z \in Z$$. And yes you are correct it is differentiable at f(z) at z=1, again I forgot to put the modulus signs around it.
Alright now that everything is cleared up, so the line z=2+ix.
Plugging this into $$|f(z)|=|(2+ix-1)(2+ix-4)^2=|(1+ix)(-x^2-4ix+4)|=|-x^2-4ix+4-ix^3+4x^2+4ix|=|-ix^3+3x^2+4|=\sqrt{(3x^2+4)^2+(x^3)^2}=$$
Then this reduces to
$$\sqrt{9x^4+24x^2+16+x^6}=\sqrt{(x^4+8x^2+16)(x^2+1)}=\sqrt{(x^2+4)^2(x^2+1)}=(x^2+4)\sqrt{x^2+1}.$$
Then the derivative of this would be:
$$2x\sqrt{x^2+1}+\frac{1}{\sqrt{x^2+1}}[x^3+4x].$$

Then setting this equal to zero and solving we get:
$$2x\sqrt{x^2+1}=-\frac{1}{\sqrt{x^2+1}}[x^3+4x],$$ giving $$2x^3+2x=-x^3-4x, x(3x^2+6)=0$$
Thus our solutions are x=0 and x=square root of (-1/2) but this is a complex number and x is a real number so this cannot happen.

Thus our solution is x=0.
Looking at the derivative again, and then taking this value for x=0 we get |f''(z)|=2 which is positive. Thus the line x=0 is a minimum.
Thus the line z=2 contains a relative minimum.

Is this correct. And therefore there are no relative maximum?

 

[HallsofIvy]

You will probably find it simpler to use the fact that $\sqrt{f(x)}$ has a minimum where f(x) itself does: specifically, the derivative of $(f(x))^{1/2}$ is $(1/2)f(x)f'(x)$ which is equal to 0 only where f(x)= 0 or f'(x)= 0. In other words, you really only need to look at the derivative of
$$x^+ 9x^4- 24x^2+16$$
which turns out to be relatively simple.

 

[wany]

Did you mean $$x^6+9x^4+24x^2+16?$$
So taking the derivative and setting this equal to zero then solving for it we get that x=0.
So the line is z=2+i(0)=2.