# Homework Help: Maximum value for a analytic function

1. Dec 27, 2009

### MrGandalf

Hi. I need someone to look at my attempt at a solution, and guide me towards the correct way to solving this. Thanks.

1. The problem statement, all variables and given/known data
Determine the maximum value of [ilatex]|3z^2 - 1|[/ilatex] in the closed disk [ilatex]|z| \leq 1[/ilatex] in the complex plane. For what values of z does the maximum occur?

The attempt at a solution
There is a theorem that says that all maximum values of an analytic function in a disc occurs at the bound, so the max values will be on some point on the circle [ilatex]|z|=1[/ilatex]. We can easily see that the maximizing points are [ilatex]z = \pm i[/ilatex], and in those cases we get [ilatex]|3(i)^2 - 1| = |-3-1| = |-4| = 4[/ilatex].

2. Dec 27, 2009

### JSuarez

It's the modulus that attains the maximum at the boundary, not the function; after all, C isn't ordered, so there is no extreme points for complex valued functions.

Regarding the solution, it's correct, but I think it needs more justification; try to interpret 3z^2 - 1 geometrically, when z is restricted to the unit circle.

3. Dec 27, 2009

### 3029298

It might be helpful to write $$z=e^{i\phi}$$ on the unit circle...

4. Dec 27, 2009

### JSuarez

You might do that, and compute the extreme points (it's not asked but the maximum modulus principle has a counterpart: the minimum modulus principle). But notice that the image of z^2, when z is in the unit circle is a also the unit circle (run over twice); multiplication by 3 will result in a circle with that radius, and -1 is a translation. You should be able to represent graphically the image of 3z^2 - 1, when z is in S1; from that, it's very easy to see where the modulus is maximized (and minimized).

5. Dec 27, 2009

### MrGandalf

I am still not sure how to show mathematically how +/- i will maximize the function.

I tried doing a geometrical approach with drawing a circle with radius 3 and center in -1, but I can't see how it becomes obvious where it is maximized... I also failed to see how looking at the exponential form would help. Maybe I'm still locked in Holiday-mode, but even after all your hints and suggestions I'm still drawing a blank on this one. :)

6. Dec 27, 2009

### JSuarez

If you have a circle with radius 3 and centre at -1, what is the point that is farthest from the origin? Will it not be the point with $$\theta = \pm\frac{\pi}{4}$$?

7. Dec 28, 2009

### MrGandalf

Ah, of course! Thanks for explaining this to me.