Find the Laurent series expansion of
f(z) = 1
z
3
(1 − z)
anyone want thanks
give 5 ❤= take 30 ❤
Answers
Answer:
12
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I assume you want the Laurent series expansion for this function at 0 (since this is the point those annuli are centered at). The standard way to do this is not to use the integral formula at all--- just the uniqueness of the Laurent series expansion (which I guess is often proved using the integral formula). You do some algebra to find series representations of this function in integer powers of z that converge in the given annuli. By uniqueness these series have to be the Laurent series.
From partial fractions
1(z−1)(z−2)=−1z−1+1z−2
so on either annulus it is enough to find the Laurent series expansions of the functions 1z−1 and 1z−2 and then combine them term by term.
If |z|>2 then note that
1z−2=1z11−2z−1
and the fact that |z|>2 implies that |2z−1|<1. If you recall the geometric series formula,
11−u=1+u+u2+⋯,|u|<1,
then putting in u=2z−1 you see that
1z−2=1z(1+(2z−1)+(2z−1)2+⋯)
holds for all |z|>2. If you just expand this out you get the Laurent series for 1z−2 in this region. Of course it is much cleaner if you use sigma notation to write it out.
On the annulus 1<|z|<2 of course the above method does not work. But we also have
1z−2=−121(1−z2)
and if 1<|z|<2 then |z2|<1 so the geometric series formula can be used again in a slightly different way here.
The Laurent series expansion of 1z−1 is easier since 1z−1=1z11−z−1 and |z−1|<1 holds for all z in either annulus. So the Laurent expansion of this function is the same in both regions.
The ideas used here generalizes to finding the Laurent series other rational functions on other annuli of course.
Explanation:
your all ans ok byby
Let’s find the Laurent series about 0, since you haven’t specified the center. Firstly, it is clear that the singularities are at z=−1 and z=−3 . These points will determine the regions where we have different Laurent Series. We can find the series for this function by taking advantage of the well known series:
11−ω=1+ω+ω2+...=∑n=0∞ωn
for |ω|<1
First rewrite the function as a sum of two fractions:
f(z)=1(z+1)(z+3)=Az+1+Bz+3=(A+B)z+(3A+B)(z+1)(z+3)
It follows that A=12 and B=−12 so that:
f(z)=12(1z+1−1z+3)
Now we are ready to find the series. First notice that as we expand a disc from 0 the function is analytic until we reach z=−1 so the function is analytic on D(0,1) . This is our first region of interest. Since the function is analytic the Laurent Series will actually just be a Taylor series.