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February 11th, 2017, 12:46 PM   #1
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Prove that |sin(z)| <= 3 and |z|<=1

How do I solve this please?
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February 11th, 2017, 12:55 PM   #2
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there has to be more to this problem than what you've written

do you mean show that

$|z|\leq 1 \Rightarrow |\sin(z)| \leq 3 ,~\forall z \in \mathbb{C}~?$
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February 11th, 2017, 01:05 PM   #3
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I'll assume you mean what Romsek said which is the only reasonable thing I can think you might. You can show this by noticing that $|e^z| = e^{\Re(z)}$ and using the expression
$$ \sin(z) = \frac{e^{iz} - e^{-iz}}{2i}.$$

I'll leave you to the details as they are straightforward.
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February 13th, 2017, 01:18 PM   #4
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Quote:
Originally Posted by romsek View Post
there has to be more to this problem than what you've written

do you mean show that

$|z|\leq 1 \Rightarrow |\sin(z)| \leq 3 ,~\forall z \in \mathbb{C}~?$
Sorry I written the question wrong.

Prove that |sin(z)| <= 3 given |z| <= 1
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