Ug, a couple more tough problems

[Wishbone]

the first one says:

Prove that:
a) $$|sin z | \geq |sin x|$$
b) $$|cos z | \geq |cos x|$$Where I guess z = x+iy...

What I have done:
Well I am pretty stumped on this one, I have though about expanding sin z into $$(sin x) (cosh y) + (i cos x) (sinh y)$$. I don't think that helps me prove anything, but it seems like more terms means it would be greater than just a sin x :uhh:

Second problemo:

We see the anuglar momentum components
$$(L_x - i L_y) != (L_x +iL_y)*$$Gosh I've tried a lot on this one, I really don't know too too much about QM, so its been tough. It just seems to go against the definition of a conjugate, so I dunno...

 

[Tide]

First:

I don't think that helps me prove anything

Think a little harder! :)

HINT: What is the absolute value of the expression?

Second: What exactly does that exclamation point mean?

 

[Wishbone]

oh i didnt know how to put the does not equal to sign, so I put the !=.

 

[Wishbone]

HINT: What is the absolute value of the expression?

hmmmm well of course I know |z| = $$\sqrt{x^2 +y^2}$$

so would |sin z| = $$\sqrt{(sin x)^2 (cosh y)^2 + (i cos x)^2 (sinh y)^2}$$

 

[Tide]

Not quite. You need to multiply sin z by its complex conjugate, i.e. the i under the radical doesn't belong there.

 

[Wishbone]

hmm ok so I get$$(sin^2 x) (cosh^2y) + (cos^2 x) (sinh^2 y)$$

but that gives me $$|sin z|^2$$, and then hmm, I need to take the sqaure root of that to get get back to |sin z|?

 

[quantumdude]

hmm ok so I get$$(sin^2 x) (cosh^2y) + (cos^2 x) (sinh^2 y)$$

but that gives me $$|sin z|^2$$, and then hmm, I need to take the sqaure root of that to get get back to |sin z|?

I think you'll find the identity $\cos^2(x)=1-\sin^2(x)$ quite helpful in finishing this off.

As for the QM question I think what they're driving at here is that you don't take the complex conjugate of operators. Instead you take the Hermitian conjugate. That means that $L_x-iL_y=(L_x+iL_y)^{\dagger}$.

And by the way the not-equal-to symbol is given by $\neq$ (click the image to see the code).