Question

Expand F(z) = (7z-2) /z(z+1)(z-2) for the region
a) 1<| z+1|< 3(b) |z+1|>3 (c) 2<|z-2|<3​

Answer 1

Answer:

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(1z+1−1z+3)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=−1z=−1 so the function is analytic on D(0,1)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. We find it by writing:

f(z

Step-by-step explanation:

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