Residue of exp(1/(z+i)): How to Find and Solve for z=-i

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In summary, the conversation discusses finding the residue of z=-i for the function exp(1/(z+i)). It is noted that the function has an essential singularity at z=-i, making the usual formula for finding the residue for a pole of order m not applicable. Instead, the function is expanded as a Laurent series around the pole and the residue is found as the coefficient in front of the term proportional to \frac{1}{z+i}. The process is briefly explained and the person asking for help is advised to expand e^x as a Taylor series and plug in x=1/(z+i).
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Hi, I want to know how you find the residue of z=-i for the function exp(1/(z+i)). Clearly, the function has an essential singularity at z=-i so the good ol' formula for the residue for a pole of order m, doesn't really work here. What do I do? :)
 
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You expand the function as a Laurent series around the pole and recall that the residue at that point is the coefficient in front of term proportional to [itex]\frac{1}{z+i} [/itex]
 
  • #3
okay, can you help me how that is done. My book's section os Laurent series is quite poor. The only thing I know about expanding functions as them, is that you can sometimes use geometric series. You don't need to say how I should do it, just hint me at where to start.
 
  • #4
Just expand e^x as a usual Taylor series, then plug in x=1/(z+i) and it's done!
 

FAQ: Residue of exp(1/(z+i)): How to Find and Solve for z=-i

What is the residue of exp(1/(z+i))?

The residue of a function at a point is the value of the coefficient of the term with the highest negative power in the Laurent series expansion around that point. In this case, the Laurent series expansion of exp(1/(z+i)) around z=-i is 1/(z+i). Therefore, the residue at z=-i is 1.

How do you find the residue of exp(1/(z+i))?

To find the residue of a function at a point, you can use the formula Res(f,z0) = lim(z→z0) (z-z0)f(z). In this case, the point z0 is -i, so the residue is Res(exp(1/(z+i)), -i) = lim(z→-i) (z+i)exp(1/(z+i)) = (1-i)exp(1/(z+i)) = 1.

Why is the residue of exp(1/(z+i)) at z=-i equal to 1?

This is because the coefficient of the term with the highest negative power in the Laurent series expansion of exp(1/(z+i)) around z=-i is 1/(z+i), which is equal to 1 when evaluated at z=-i. Therefore, the residue at z=-i is 1.

Can the residue of exp(1/(z+i)) at z=-i be negative?

No, the residue of a function at a point cannot be negative. In this case, the residue is equal to 1, which is a positive value.

Is the residue of exp(1/(z+i)) at z=-i affected by the function's poles and zeros?

Yes, the residue at a point can be affected by the presence of poles and zeros in the function. However, in this case, exp(1/(z+i)) has a simple pole at z=-i, so the residue is not affected by any other poles or zeros.

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