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

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In summary, in this conversation, the speaker is discussing how to find poles and evaluate them for a given function. They provide an equation for finding the residue at a simple pole and demonstrate using an example. They then ask for clarification on a step in the solution and it is explained that it involves using the trigonometric identity for sine.
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laura_a
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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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  • #2
For simple pole at [tex]z_0[/tex], the residue of f(z) is given by
[tex]\text{Res}\left[f(z), z_0\right]= \lim_{z\rightarrow z_0} (z-z_0) f(z)[/tex]

eg.
[tex]\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}[/tex]

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
[tex]\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}[/tex]
 

1. How do I solve the equation sin(z)/2i = 1/4(e - 1/e) if it has poles?

The first step is to rewrite the equation in terms of complex numbers. This will give us sin(z)/2i = (e - e^(-z))/(4i). Then, we can multiply both sides by 4i to eliminate the denominator. This will give us sin(z) = e - e^(-z). From here, we can use the property of complex exponential functions that e^(iz) = cos(z) + isin(z). Therefore, we can rewrite the equation as sin(z) = e - (cos(z) - isin(z)). This can be further simplified to sin(z) = e - cos(z) + isin(z). Finally, we can equate the real and imaginary parts of both sides to solve for z.

2. Can I use the quadratic formula to solve sin(z)/2i = 1/4(e - 1/e) if it has poles?

No, the quadratic formula is only applicable to equations with the form ax^2 + bx + c = 0. The equation sin(z)/2i = 1/4(e - 1/e) does not have this form and therefore cannot be solved using the quadratic formula.

3. How can I determine the number of poles in the equation sin(z)/2i = 1/4(e - 1/e)?

The number of poles in this equation can be determined by analyzing the behavior of the denominator, which is (e - 1/e). Since this expression contains an exponential term, it will have poles at the points where e^z = 0, which are located at z = iπ/2 and z = -iπ/2. Therefore, the equation has two poles.

4. Can I use the residue theorem to solve sin(z)/2i = 1/4(e - 1/e) if it has poles?

Yes, the residue theorem can be used to solve this equation. The residue theorem states that the integral of a function around a closed contour is equal to the sum of the residues of the function at its singular points inside the contour. In this case, the singular points are the poles at z = iπ/2 and z = -iπ/2. By finding the residues at these points and applying the theorem, the equation can be solved.

5. Can I use the complex logarithm to solve sin(z)/2i = 1/4(e - 1/e) if it has poles?

Yes, the complex logarithm can be used to solve this equation. The complex logarithm is defined as ln(z) = ln|z| + iArg(z), where Arg(z) is the argument of z in the complex plane. By taking the logarithm of both sides of the equation sin(z)/2i = 1/4(e - 1/e), we can simplify the equation and solve for z. However, it is important to note that the complex logarithm is a multi-valued function and therefore may produce more than one solution for z.

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