# Question regarding maximum on a unit disc

1. Jun 8, 2010

### librastar

1. The problem statement, all variables and given/known data

Find the maximum of |ez| on the closed unit disc.

2. Relevant equations

|ez| is the modulus of ez
z belongs to complex plane
Maximum Madulus Theorem - Let G be a bounded open set in complex plane and suppose f is a continuous function on G closure which is analytic in G. Then max{|f(z)|: z in G closure} = max{|f(z)|: z in boundary of G}

3. The attempt at a solution

At first glance I was thinking about using Maximum Modulus Theorem by setting f(z) = ez and let G be a open unit disc. Clearly f(z) is analytic in G and is continuous on G closure (the closed unit disc), so I should be able to apply MMT to conclude that the modulus assumes maximum on the boundary.
However I got confused when determining how do I describe |f(z)| on the boundary, in another word, how do I determine what value of z to use?

Please let me know if my approach is right and perhaps give me some advice or hint.
I appreciate all the helps you provide.

2. Jun 8, 2010

### lanedance

as on the boundary |z| = 1, how about starting by writing it in the form z = 1. e^(i theta)

3. Jun 8, 2010

### Coto

Since z lies on the unit circle, we can write it as: max{|f(z)|: z in boundary of G} = max{|f(z)|: z = exp($$i\theta$$), $$\theta \in [-\pi,\pi)$$}

However, what do you know about |$$e^z$$| when $$z = e^{i\theta}$$?

4. Jun 8, 2010

### librastar

I'm not sure where this is going, in this case we will just have exp(exp($$i\theta$$))?

I also know that e^(i$$\theta$$) = cos($$\theta$$) + i*sin($$\theta$$), but I don't know if this will help my solution or not.

5. Jun 8, 2010

### Coto

It does. Calculate explicitly the modulus of $$e^z$$ using the identity you provided above for $$e^{i\theta}$$.

6. Jun 8, 2010

### librastar

So now I have |ecos($$\theta$$)+isin($$\theta$$)|

|ecos($$\theta$$)*eisin($$\theta$$)|
= |ecos($$\theta$$)*(cos(sin($$\theta$$))+isin(sin($$\theta$$)))|

Now I don't know why I got stuck here....

7. Jun 8, 2010

### lanedance

not too sure what you did there in the last step... but you need to remember the complex conjugate in the magnitude... try that before you simplify
$$|e^{ cos(\theta)}e^{i sin(\theta)}|^2 = (e^{ cos(\theta)}e^{i sin(\theta)})(e^{ cos(\theta)}e^{i sin(\theta)})^*$$

8. Jun 8, 2010

### librastar

thanks for the tip.

Well, I know that (ez)* = e$$z^*$$

So I can simply $$|e^{ cos(\theta)}e^{i sin(\theta)}|^2 = (e^{ cos(\theta)}e^{i sin(\theta)})(e^{ cos(\theta)}e^{i sin(\theta)})^*$$ to:

$$(e^{ cos(\theta)}e^{i sin(\theta)})(e^{ cos(\theta)}e^{-i sin(\theta)})$$

Then by combing the exponential functions with the same exponent, then I obtain:

$$(e^{ 2cos(\theta)})$$

So after taking the square root of the whole thing,

$$|e^z| = e^ { cos(\theta)}$$

The max value of $$cos(\theta)$$ = 1

So the maximum of $$|e^z| = e$$?

9. Jun 8, 2010

### lanedance

that looks good to me

10. Jun 8, 2010

Thanks.