Finding Residues of a Function

• asi123
In summary, the function f(z) = cos(1/(z-2)) has an essential singularity at z=2, meaning there is no residue. However, the residue can still be found by finding the Laurent series of the function. In the case of the function cos(1/(z-2)), the residue is 0 if the power of z-2 is always even and 1 if the power of z-2 is odd.
asi123
residues 2 :)

Homework Statement

Ok, so I have this function and I need to find the residues for it.
I developed the function into Laurent series and my question is, how can I find a(-1) which is the residues?

The Attempt at a Solution

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The function is
$$f(z)= cos(\frac{1}{z-2})$$
and you correctly used the McLaurin series for cos(u), with u= 1/(z-2), to write the Laurent series
$$\sum_{n= 0}^\infty \frac{(-1)^n}{(2n)!} (z-2)^{-2n}$$
Since the power of z- 2 is always even, there is no (z-2)-1 term: the coefficient is 0.

HOWEVER, because that series has an infinite number of negative powers with non-zero coefficient, z= 2 is an essential singularity, not a pole. There is NO residue.

HallsofIvy said:
The function is
$$f(z)= cos(\frac{1}{z-2})$$
and you correctly used the McLaurin series for cos(u), with u= 1/(z-2), to write the Laurent series
$$\sum_{n= 0}^\infty \frac{(-1)^n}{(2n)!} (z-2)^{-2n}$$
Since the power of z- 2 is always even, there is no (z-2)-1 term: the coefficient is 0.

HOWEVER, because that series has an infinite number of negative powers with non-zero coefficient, z= 2 is an essential singularity, not a pole. There is NO residue.

Oh, so an essential singularity has no residue?

Thanks.

Thanks again.

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HallsofIvy said:
The function is
$$f(z)= cos(\frac{1}{z-2})$$
and you correctly used the McLaurin series for cos(u), with u= 1/(z-2), to write the Laurent series
$$\sum_{n= 0}^\infty \frac{(-1)^n}{(2n)!} (z-2)^{-2n}$$
Since the power of z- 2 is always even, there is no (z-2)-1 term: the coefficient is 0.

HOWEVER, because that series has an infinite number of negative powers with non-zero coefficient, z= 2 is an essential singularity, not a pole. There is NO residue.

I am informed that, according to Wikipedia, I am wrong about this. That is, that the residue is still the coefficient of the -1 power in the Laurent series. The difference is just that the formula for the residue, taking the nth derivative where n is the order of the pole, cannot be used but you still can find it by finding the Laurent series itself. In the case of your first problem, the residue is 0 and for your new problem, it is 1.

1) What are residues of a function?

Residues of a function are the complex numbers that are left over when the function is evaluated at a point where it is not defined.

2) How do you find residues of a function?

To find the residues of a function, you can use the residue theorem, which states that the residue of a function at a point is equal to the coefficient of the (1/z) term in its Laurent series expansion at that point.

3) Why are residues important in complex analysis?

Residues are important in complex analysis because they allow us to evaluate complex integrals and calculate the values of certain functions at points where they are not defined. They also play a crucial role in the study of meromorphic functions and the behavior of functions near their singularities.

4) What are some applications of finding residues of a function?

Finding residues of a function has many applications in mathematics and physics. It can be used to solve complex integrals, evaluate trigonometric and logarithmic functions, and calculate residues at poles of a function. In physics, residues are used in the study of quantum mechanics, electromagnetism, and fluid dynamics.

5) Is it possible to find residues of a function graphically?

No, it is not possible to find residues of a function graphically. Residues are calculated algebraically using the residue theorem or other methods such as Cauchy's integral formula. However, graphical representations can be used to visualize the behavior of a function and its residues near singularities.

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