expend f(z)=1/(z+1)(z+3) in a lourent's series valid in
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Can you expand f(z)=1/(z+1)(z+3) by the Laurent series valid region?
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2 Answers
Eric Neville
Answered December 31, 2018
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=−1z=−1 and z=−3z=−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∞ωn11−ω=1+ω+ω2+...=∑n=0∞ωn
for |ω|<1|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)f(z)=1(z+1)(z+3)=Az+1+Bz+3=(A+B)z+(3A+B)(z+1)(z+3)
It follows that A=12A=12 and B=−12B=−12 so that:
f(z)=12(
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this is your answer, please show the picture.
