How can I find the Laurent series expression for z z 2 1 )?

As z^2 + 1 = (z-i)(z+i) you can multiply your function by one of the factors, find the Taylor series, then divide all terms by that factor. And voila! there’s a Laurent series.

How is residue of Laurent series calculated?

The residue Res(f, c) of f at c is the coefficient a−1 of (z − c)−1 in the Laurent series expansion of f around c. Various methods exist for calculating this value, and the choice of which method to use depends on the function in question, and on the nature of the singularity.

What is the principal part of a Laurent expansion?

The portion of the series with negative powers of z – z 0 is called the principal part of the expansion. It is important to realize that if a function has several ingularities at different distances from the expansion point , there will be several annular regions, each with its own Laurent expansion about .

What are poles and residue?

The different types of singularity of a complex function f(z) are discussed and the definition of a residue at a pole is given. The residue theorem is used to evaluate contour integrals where the only singularities of f(z) inside the contour are poles.

Is Laurent expansion unique?

expansion of a function f(z) ⁢ in an annulus r<|z−z0|. i.e. aν=bν a ν = b ν , for any integer ν , whence both expansions are identical. …

How to find all Laurent series centered at z = 0?

Problem Find all Laurent series centered at z = 0 for f ( z) = 1 z ( z − 1) ( z − 2) . In each region, the Laurent expansion is obtained by expnading the terms in powers of z. Obviously the first term is already in that form. Region I. − 1 z − 1 = 1 1 − z.

How do you find the Laurent expansion for each region?

In each region, the Laurent expansion is obtained by expnading the terms in powers of z. Obviously the first term is already in that form. Region I. − 1 z − 1 = 1 1 − z. Since | z | < 1, we may expand this term using an appropriate geometric series: 1 1 − z = ∑ k = 0 ∞ z k. Similarly, which converges in | z | < 2.

What is the Laurent series expansion of a singularity?

The Laurent series expansion is defined on a “deleted neighborhood” around a singularity, in this case, {z: 0 < | z − 0 | < R}. In this deleted neighborhood, e1 / z is analytic. So for any point in this neighbourhood, we can expand ez first and then substitute 1 / z in. As you can see in the Laurent expansion you gave,…

Is there an integral formula for the Laurent series?

What is it that doesn’t seem possible? There is a general integral formula for Laurent seriescoefficients, but they are instead often found using some other known series, like in this case, where $f(z) = a_0 + a_1z +a_2z^2+\\cdots$ is valid for all $z\\in\\mathbb C$, and it follows that $f(1/z)= a_0 + a_1/z +a_2/z^2+\\cdots$ is valid for $z eq 0$.