Question regarding maximum on a unit disc

[librastar]

Homework Statement

Find the maximum of |e^z| on the closed unit disc.

Homework Equations

|e^z| is the modulus of e^z
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}

The Attempt at a Solution

At first glance I was thinking about using Maximum Modulus Theorem by setting f(z) = e^z 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.

 

[lanedance]

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

 

[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 $$z = e^{i\theta}$$?

 

[librastar]

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 $$z = e^{i\theta}$$?

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.

 

[Coto]

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

 

[librastar]

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

So now I have |e^{cos($$\theta$$)+isin($$\theta$$)}|

|e^{cos($$\theta$$)}*e^{isin($$\theta$$)}|
= |e^{cos($$\theta$$)}*(cos(sin($$\theta$$))+isin(sin($$\theta$$)))|

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

 

[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
$$|e^{ cos(\theta)}e^{i sin(\theta)}|^2 = (e^{ cos(\theta)}e^{i sin(\theta)})(e^{ cos(\theta)}e^{i sin(\theta)})^*$$

 

[librastar]

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
$$|e^{ cos(\theta)}e^{i sin(\theta)}|^2 = (e^{ cos(\theta)}e^{i sin(\theta)})(e^{ cos(\theta)}e^{i sin(\theta)})^*$$

thanks for the tip.

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

 

[lanedance]

that looks good to me

 

[librastar]

that looks good to me

Thanks.