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February 11th, 2017, 11:46 AM  #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, 11:55 AM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,397 Thanks: 709 
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, 12:05 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 114 Thanks: 45 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, 12:18 PM  #4 
Newbie Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0  

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