How do i get sin(z)/2i = 1/4(e - 1/e) ? Resides at Poles question?

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The discussion focuses on finding the residue of the function f(z) = sin(z)/(z^2 + 1) at its poles. The user successfully identifies the simple pole at z = i and calculates the residue as sin(i)/2i. To derive the expression 1/4(e - 1/e), the user is guided to use the identity sin(x) = (e^(ix) - e^(-ix))/(2i). This transformation allows for the simplification of sin(i) to yield the desired result. Understanding this relationship between sine and exponential functions is crucial for completing the problem.
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Homework Statement



Find Poles & eval resides at those points for

f(z) = sin(z)/(z^2+1)



Homework Equations



I'm using Res f(z) = [g^m-1 (z_0)] / (m-1)!

The Attempt at a Solution



So I worked out the f(z) = 1/(z-i) so g(z) = sin(z) /(z+i) for my first one did the sub for z=i and got sin(i)/2i

But the working out (i'm working on past papers so I have the worked solutions, just not very detailed) The next line shows;

= 1/4(e-(1/e))

What one Earth don't I know about sin that gets to this point??
 
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For simple pole at z_0, the residue of f(z) is given by
\text{Res}\left[f(z), z_0\right]= \lim_{z\rightarrow z_0} (z-z_0) f(z)

eg.
\text{Res}\left[\frac{1}{(z^2+1)}, i \right]= \lim_{z\rightarrow i} \frac{z-i}{z^2+1} =\lim_{z\rightarrow i} \frac{1}{z+i}=\frac{1}{2i}

now, looks like you have got it right so far with sin(i)/2i, all it is need to get that next line is just by using
\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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