Is this the answer for what is the residue of f(z)=e^z/[(z-1)^2(z+2)]

  • Thread starter laura_a
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In summary: Rightarrow g'\left( 1 \right) = \frac{e}{{9}}In summary, the poles of f(z) are located at z=1 (of order 2) and z=-2 (of order 1). The residue at z=1 is -e/9, and the residue at z=-2 is 1/(9e^2). It is important to double check the calculation for the residue at z=1.
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laura_a
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Homework Statement


This was an exam question I had today:

Give the location and order of each pole of
f(z)=e^z/[(z-1)^2(z+2)]
and evaluate the reside at these points


Homework Equations



Res(f(z)) = g(p-1)(a) / (p-1)!



The Attempt at a Solution



g(z) = e^z/(z+2)
g'(z) = -e^z/(z+2)^2
let a=1
g'(1) = -e^1/9

So res(f,1) = -e/9

Then I did
g(z) = e^z/(z-1)^2
(this is the part I think I might have stuffed up, because I did use the formula above but I had m=1 for f(z) which was 1/(z+2) so I thought it was right? anyways I got

g(-2) = e^(-2) / (-3)^2 = res(f,-2)=1/(9e^2)

It would mean a lot to me if someone could let me know because I am trying to work out if I should be depressed because I failed or if I can be happy because I passed :)
 
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  • #2
At first glance I agree with your first answer for res(f,-2) but not for res(f,1).

[tex]
g\left( z \right) = \frac{{e^z }}{{\left( {z + 2} \right)}} \Rightarrow g'\left( z \right) = \frac{{e^z }}{{\left( {z + 2} \right)}} - \frac{{e^z }}{{\left( {z + 2} \right)^2 }}
[/tex]
 

1. What is the residue of f(z) at z = 1?

The residue of f(z) at z = 1 is 1/4.

2. How do you calculate the residue of a complex function?

To calculate the residue of a complex function, you first need to find the poles of the function. Then, you can use the formula Res(f,z0) = lim(z->z0) [(z-z0)f(z)].

3. What does the residue of a function represent?

The residue of a function at a specific point represents the coefficient of the (z-z0)^-1 term in the Laurent series expansion of the function around that point.

4. Can the residue of a function be negative?

Yes, the residue of a function can be negative. The sign of the residue depends on the direction of integration around the pole.

5. How is the residue of a function related to its singularities?

The residue of a function is directly related to its singularities. Specifically, the residue at a pole z0 is equal to the coefficient of the (z-z0)^-1 term in the function's Laurent series expansion around that pole.

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