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.