Solve ML Inequality: Establishing Inequality for Complex Function Modulus

[shebbbbo]

Having trouble with this question:


The question is: establish the inequality

|$\int$e^izzdz| $\leq$ $\pi$(1-e^-R2)/4R

on C {z(t) = Re^it, t $\in$ [0,$\pi$/4, R>0


When i saw the modulus of an integral i thought ML inequality.

I think the length will be R$\pi$/4 but I am struggling with finding the maximum of e^izz. I tried changing to e^{i(r2(cos(2t)+isin(2t)}. but i don't feel any closer to the result.

Am i on the right track, and can anyone help me with finding the max of the function.

thanks

 

[jackmell]

Yes, you're on the right track. Convert it to sines and cosines:

$$R\int_{0}^{\pi/4} e^{-R^2 \sin(2t)+iR^2 \cos(2t)} ie^{it}dt$$

and the absolute value of that is less than:

$$R\int_0^{\pi/4} e^{-R^2 \sin(2t)}dt$$

now, draw the plot of sin(2x) over the range of 0 and pi/4 and can you see that:

$$\left|e^{-R^2\sin(2t)}\right|<e^{-R^2(4t/\pi)}$$

Draw the line $y=\frac{4}{\pi}t$ over sin(2t) to see that.

 

[shebbbbo]

im now more confused...

are you saying i don't need the ML inequality?

I don't understand where the R in the first line comes from? and where do you go from the last line to get closer to the solution?

im sorry i havnt understood

 

[jackmell]

Been out. I'm just placing an upper bound on the integrand. The R comes from letting $z=Re^{it}$ so when you substitute that into the integral, you get:

$$\int_{0}^{\pi/4} e^{i R^2 e^{2it}} Ri e^{it} dt$$

right?

Now, convert that all to sines and cosines to get:

$$Ri\int_{0}^{\pi/4} e^{iR^2(cos 2t+i\sin 2t)}e^{it}dt$$

now the absolute value of that:

$$\biggr|Ri\int_{0}^{\pi/4} e^{iR^2(cos 2t+i\sin 2t)}e^{it}dt\biggr|\leq R\int_0^{\pi/4} e^{-R^2 \sin(2t)}dt$$

right? Now, that's when we place an upper bound on the integrand by using the line $y=4/\pi t$

Draw that line and $\sin(2t)$ to see that it's underneath the sine function so that we can write:

$$\biggr|Ri\int_{0}^{\pi/4} e^{iR^2(cos 2t+i\sin 2t)}e^{it}dt\biggr|\leq R\int_0^{\pi/4} e^{-R^2 \sin(2t)}dt\leq\biggr|\int_0^{\pi/4} e^{-R^2(4t/\pi)}dt\biggr|$$

and that you can integrate directly.

 

[nugget]

now the absolute value of that:

$$\biggr|Ri\int_{0}^{\pi/4} e^{iR^2(cos 2t+i\sin 2t)}e^{it}dt\biggr|\leq R\int_0^{\pi/4} e^{-R^2 \sin(2t)}dt$$

right?

Hey Jackmell,

could you please explain how you would show that the magnitudes of some of the things in this integral are equal to one? i understand for e^it but not for e^iR2cos(2t)

 

[jackmell]

You know:

$$|e^{iv}|=1$$

for any v right? Ok, then there you go:

$$\left|e^{iR^2(\cos(2t)+i\sin(2t)}\right|=\left|e^{-R^2\sin(2t)+iR^2\cos(2t)}\right|$$

and the absolute value of that i-part is going to be one right?

$$=\left|e^{-R^2 \sin(2t)\left(\cos(R^2\cos(2t))+i\sin(R^2 \cos(2t))\right)}\right|$$

$$=\left|e^{-R^2 \sin(2t)}\right|$$