My Math Forum Prove that |sin(z)| <= 3 and |z|<=1

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 February 11th, 2017, 12:46 PM #1 Newbie   Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0 Prove that |sin(z)| <= 3 and |z|<=1 How do I solve this please?
 February 11th, 2017, 12:55 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,753 Thanks: 896 there has to be more to this problem than what you've written do you mean show that $|z|\leq 1 \Rightarrow |\sin(z)| \leq 3 ,~\forall z \in \mathbb{C}~?$
 February 11th, 2017, 01:05 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 274 Thanks: 141 Math Focus: Dynamical systems, analytic function theory, numerics 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.
February 13th, 2017, 01:18 PM   #4
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 Originally Posted by romsek 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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