Proving Inequality: Solving Im(z) and Re(z) with Triangle Inequality

[Cosmossos]

Hello
I need to prove this inequality:
http://img6.imageshack.us/img6/2047/unledwp.jpg

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where y=im(z) ,x=Re(z).

I used the triangle inequality but I got stuck.
Can someone show me how to do it? specially the left side of the inequality.
thanks

 

[SammyS]

Show what you have tried, and where you got stuck.

That way it will be easier for use to give the appropriate help.

 

[sir_manning]

Don't forget about the reverse triangle inequality: |x - y| >= ||x| - |y||

 

[Cosmossos]

Here:
http://img641.imageshack.us/img641/5589/unlediny.jpg

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[sir_manning]

It will be easier to break this into two problems. First prove that $\left| sin z \right| \leq \frac{e^{y} + e^{-y}}{2}$ , then prove that $\left| sin z \right| \geq \frac{e^{\left| y \right|} + e^{- \left| y \right|}}{2}$.

Also, use $\left| sin z \right| = \frac{e^{i(x+iy)} - e^{-i(x+iy)}}{2i}$ .

 

[Cosmossos]

It will be easier to break this into two problems. First prove that $\left| sin z \right| \leq \frac{e^{y} + e^{-y}}{2}$ , then prove that $\left| sin z \right| \geq \frac{e^{\left| y \right|} + e^{- \left| y \right|}}{2}$.

Also, use $\left| sin z \right| = \frac{e^{i(x+iy)} - e^{-i(x+iy)}}{2i}$ .

That's what i did. can you please look at my answer? isn't it correct?
thank you.

 

[sir_manning]

Sorry, I see that you did write $\left| sin z \right| = \frac{e^{i(x+iy)} - e^{-i(x+iy)}}{2i}$. However, I don't understand how you came up with your answer: where did the absolute value signs in $\left| \frac{e^{y} + e^{-y}}{2} \right|$ emerge from? You can't just insert them. And how did you re-arrange the inequality? Was there a typo in your original statement of the problem? In any case, your answer doesn't prove the inequality, because I cannot see its validity just by looking at it. With these types of problems, you really need to break it down to something like $-e^{-y} \leq e^{-y} \; \Rightarrow \; -1 \leq 1$, which we can all agree is true. Also, in proofs you *need* to show your steps, and you always should here anyways so we can help you out.

Alright, let's try doing this one part at a time. First, prove that:

$\left| sin z \right| \leq \frac{e^{y} + e^{-y}}{2}$, or

$\left| \frac{e^{i(x+iy)} - e^{-i(x+iy)}}{2i} \right| \leq \frac{e^{y} + e^{-y}}{2}$. Cancel the 2's, multiply by i/i and rearrange exponentials on the left,

$\left| -i e^{ix} e^{-y} + i e^{-ix} e^y \right| \leq e^{y} + e^{-y}$ ...Now try applying the triangle inequality to this. After proving this, a similar approach is used for $\left| sin z \right| \geq \frac{e^{\left| y \right|} + e^{- \left| y \right|}}{2}$