# Complex Integration Questions and Answers – Laurent Series

This set of Complex Integration Multiple Choice Questions & Answers (MCQs) focuses on ” Laurent Series”.

1. Which of the following is obtained by evaluating $$f(z)= \frac{z^2-1}{z^2+5z+6}$$ valid in the region 2<|z|<3 as a Laurent’s Series?
a) $$1+ ∑_{n=0}^∞(-1)^n (\frac{z}{2}) – \frac{8}{3} ∑_{n=0}^∞(-1)^n(\frac{z}{3})$$
b) $$1+(\frac{3}{z}) ∑_{n=0}^∞(-1)^n (\frac{2}{z})^n-\frac{8}{3} ∑_{n=0}^∞(-1)^n(\frac{z}{3})^n$$
c) $$1+(\frac{3}{z}) ∑_{n=0}^∞(-1)^n (\frac{2}{z})^n-\frac{8}{z} ∑_{n=0}^∞(-1)^n(\frac{3}{2})^n$$
d) $$1+(\frac{3}{2}) ∑_{n=0}^∞(-1)^n (\frac{2}{3})^n-\frac{8}{z} ∑_{n=0}^∞(-1)^n (\frac{3}{z})^n$$

Explanation:
$$f(z)= \frac{(z^2-1)}{(z^2+ 5z+6)}=\frac{(z^2-1)}{(z+2)(z+3)}$$
$$\frac{(z^2-1)}{(z^2+ 5z+6)}=1- \frac{(5z+7)}{(z+2)(z+3)}$$………….. (1)
$$\frac{(5z+7)}{(z+2)(z+3)}= \frac{ A}{z+2}+ \frac{B}{z+3}$$………………. (2)
5z+7 = A(z+3) + B(z+2)
Solving by Partial fractions, we get
$$\frac{(5z+7)}{(z+2)(z+3)}= \frac{(-3)}{(z+2)}+ \frac{8}{(z+3)}$$
Substituting in (1)
$$f(z)=\frac{(z^2-1)}{(z^2+ 5z+6)}=1+ \frac{3}{(z+2)}- \frac{8}{(z+3)}$$ …………………… (3)
Given that 2<|z|<3 2 < |z| and |z| < 3
$$|\frac{2}{z}|$$< 1 and $$|\frac{z}{3}|$$<1
$$f(z)= 1+ \frac{3}{z} [\frac{1}{1+\frac{2}{z}}]- \frac{8}{3} [\frac{1}{1+\frac{z}{3}}]$$
$$f(z)= 1+\frac{3}{z} [1+ \frac{2}{z}]^{-1}-\frac{8}{3}[1+ \frac{z}{3}]^{-1}$$
$$f(z)=1+(\frac{3}{z}) ∑_{n=0}^∞(-1)^n (\frac{2}{z})^n-\frac{8}{3} ∑_{n=0}^∞(-1)^n (\frac{z}{3})^n$$

2. Which of the following is obtained by evaluating the function $$\frac{z-1}{(z+2)(z+3)},$$ valid in the region 2<|z|<3 by Laurent’s Series?
a) $$1+ ∑_{n=0}^∞(-1)^n (\frac{z}{2})- \frac{8}{3} ∑_{n=0}^∞(-1)^n (\frac{z}{3})$$
b) $$1+(\frac{3}{z}) ∑_{n=0}^∞(-1)^n (\frac{2}{z})^n-\frac{8}{3} ∑_{n=0}^∞(-1)^n (\frac{z}{3})^n$$
c) $$-\frac{3}{z} ∑_{n=0}^∞(-1)^n (\frac{2}{z})^n+ \frac{4}{3} ∑_{n=0}^∞(-1)^n(\frac{z}{3})^n$$
d) $$-\frac{3}{2} ∑_{n=0}^∞(-1)^n (\frac{2}{3})^n+ \frac{4}{5} ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n$$

Explanation:
Let $$f(z)= \frac{(z-1)}{(z+2)(z+3)}$$
$$\frac{(z-1)}{(z+2)(z+3)}=\frac{A}{(z+2)}+\frac{B}{(z+3)}$$
z-1=A(z+3)+ B(z+2)
Solving by Partial Fractions,
$$f(z)= -\frac{3}{z+2}+ \frac{4}{z+3}$$
Given region is 2 < |z| < 3
i.e. 2 < |z| and |z| < 3.
$$|\frac{2}{z}|$$< and $$|\frac{z}{3}|$$<1
$$f(z)= \frac{-3}{z[1+\frac{2}{z}]} + \frac{4}{3[1+\frac{z}{3}]} = \frac{-3}{z} [1+\frac{2}{z}]^{-1}+ \frac{4}{3}[1+\frac{z}{3}]^{-1}$$
$$f(z)= \frac{-3}{z} ∑_{n=0}^∞(-1)^n (\frac{2}{z})^n+ \frac{4}{3}∑_{n=0}^∞(-1)^n (\frac{z}{3})^n$$

3. Which of the following is obtained by evaluating the function $$\frac{1}{(z+1)(z+3)}$$ in Laurent’s Series valid for the region |z|>3 ?
a) $$\frac{1}{2z} ∑_{n=0}^∞(-1)^n (\frac{1}{z})^n -\frac{1}{2z} ∑_{n=0}^∞(-1)^n (\frac{3}{z})^n$$
b) $$\frac{1}{3z} ∑_{n=0}^∞(-1)^n (\frac{2}{z})^n-\frac{1}{5} ∑_{n=0}^∞(-1)^n(\frac{z}{2})^n$$
c) $$\frac{1}{z}∑_{n=0}^∞(-1)^n (\frac{1}{z})^n-\frac{1}{z} ∑_{n=0}^∞(-1)^n (\frac{z}{3})^n$$
d) $$\frac{1}{2} ∑_{n=0}^∞(-1)^n (\frac{1}{z})^n-\frac{1}{4} ∑_{n=0}^∞(-1)^n (\frac{z}{3})^n$$

Explanation:
Given: $$f(z)= \frac{1}{(z+1)(z+3)} =\frac{A}{(z+1)}+ \frac{B}{(z+3)}$$
1=A(z+3)+ B (z+1)
Solving by Partial Fractions,
$$f(z)= \frac{1}{2} (\frac{1}{z+1})-\frac{1}{2} (\frac{1}{(z+3)})$$…………… (2)
Given Region is |z|>3
3 < |z|, $$|\frac{3}{z}|$$ < 1, $$|\frac{1}{z}|$$ < 1
From (2), $$f(z)= \frac{1}{2} \frac{1}{z[1+\frac{1}{z}]}- \frac{1}{2} \frac{1}{z[1+\frac{3}{z}]}$$
$$f(z)= \frac{1}{2z}[1+\frac{1}{z}]^{-1} -\frac{1}{2z} [1+ \frac{3}{z}]^{-1}$$
$$f(z)= \frac{1}{2z} ∑_{n=0}^∞(-1)^n (\frac{1}{z})^n-\frac{1}{2z} ∑_{n=0}^∞(-1)^n (\frac{3}{z})^n$$

4. Which of the following regions is valid by evaluating $$\frac{1}{z(z-1)}$$ as Laurent’s Series in powers of z?
a) 0<|z|<1
b) |z|>3
c) 1<|z|<3
d) 2<|z|<3

Explanation:
Given $$f (z) is \frac{1}{z(z-1)}$$
f(z) is not analytic at z=0 and z=1
But, it is analytic in the following regions
I. 0 <|z|<1
II. |z|>1

$$\frac{1}{z(z-1)} = \frac{-1}{z(1-z)} = \frac{-1}{z}(1-z)^{-1}$$
$$= \frac{-1}{z}[1+z+z^2+⋯]$$
$$= -[\frac{1}{z}+1+z+z^2+z^3+⋯]$$
The region of validity of this expansion is 0 < |z| < 1.

5. Which of the following is obtained by evaluating $$\frac{z}{(z^2-3z+2)}$$ as a Laurent’s Series in the region |z|<1?
a) $$∑_{n=0}^∞z^n – ∑_{n=0}^∞(\frac{z}{2})^n$$
b) $$∑_{n=0}^∞(\frac{1}{z})^n – ∑_{n=0}^∞(\frac{z}{2})^n$$
c) $$\frac{-1}{z} ∑_{n=0}^∞(\frac{1}{z})^n + \frac{2}{z}∑_{n=0}^∞(\frac{2}{z})^n$$
d) $$\frac{1}{z-1} – 2∑_{n=0}^∞(z-1)^n$$

Explanation:
$$f(z) = \frac{z}{z^2-3z+2} = \frac{z}{(z-1)(z-2)} = \frac{A}{z-1} + \frac{B}{z-2}$$
z = A(z-2) + B(z-1)
Solving by Partial Fractions,
$$f(z)= \frac{-1}{z-1} + \frac{2}{z-2} = \frac{1}{1-z} + \frac{2}{z-2}$$
Given region is |z|<1 |z/2|<1
$$f(z) = \frac{1}{1-z} – \frac{1}{1-\frac{z}{2}} = (1-z)^{-1} – (1-\frac{z}{2})^{-1} = ∑_{n=0}^∞z^n – ∑_{n=0}^∞(\frac{z}{2})^n$$

6. Which of the following is obtained by evaluating $$f(z)= \frac{z^2-1}{z^2+5z+6}$$ valid in the region |z|<2 using Laurent’s Series?
a) $$1+∑_{n=0}^∞(-1)^n (\frac{z}{2})- \frac{8}{3} ∑_{n=0}^∞(-1)^n (\frac{z}{3})$$
b) $$1+(\frac{3}{2}) ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n-\frac{8}{3} ∑_{n=0}^∞(-1)^n (\frac{z}{3})^n$$
c) $$1+(\frac{3}{z} ∑_{n=0}^∞(-1)^n (\frac{2}{z})^n- \frac{8}{z} ∑_{n=0}^∞(-1)^n(\frac{3}{2})^n$$
d) $$1+(\frac{3}{2}) ∑_{n=0}^∞(-1)^n (\frac{2}{3})^n- \frac{8}{z} ∑_{n=0}^∞(-1)^n (\frac{3}{z})^n$$

Explanation:
$$f(z)= \frac{z^2-1}{z^2+ 5z+6}= \frac{z^2-1}{(z+2)(z+3)}$$
$$\frac{z^2-1}{z^2+ 5z+6}= 1- \frac{5z+7}{(z+2)(z+3)}$$ ………….. (1)
$$\frac{5z+7}{(z+2)(z+3)}= \frac{A}{z+2} + \frac{B}{z+3}$$ ………………. (2)
5z+7=A(z+3)+B(z+2)
Solving by Partial fractions, we get
$$\frac{5z+7}{(z+2)(z+3)}= \frac{-3}{z+2}+ \frac{8}{z+3}$$
Substituting in (1)
$$f(z)= \frac{z^2-1}{z^2+ 5z+6}= 1+ \frac{3}{z+2} – \frac{8}{z+3}$$ …………………… (3)
Given region is |z| <2
$$|\frac{z}{2}|$$<1,$$|\frac{z}{3}|<1$$
From (3)
$$f(z)= 1+ \frac{3}{2} [\frac{1}{1+\frac{z}{2}}]- \frac{8}{3}[\frac{1}{1+ \frac{z}{3}}]$$
$$= 1+ \frac{3}{2} [1+ \frac{z}{2}]^{-1} – \frac{8}{3}[1+ \frac{z}{3}]^{-1}$$
$$f(z)= 1+ (\frac{3}{2}) ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n- \frac{8}{3} ∑_{n=0}^∞(-1)^n (\frac{z}{3})^n$$

7. Which of the following is obtained by evaluating $$\frac{1}{(z+1)(z+3)}$$ in Laurent’s Series valid for the region 1 < |z| < 3 ?
a) $$\frac{1}{2z}∑_{n=0}^∞(-1)^n (\frac{1}{z})^n- \frac{1}{2z} ∑_{n=0}^∞(-1)^n (\frac{3}{z})^n$$
b) $$\frac{1}{3z} ∑_{n=0}^∞(-1)^n (\frac{2}{z})^n-\frac{1}{5} ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n$$
c) $$\frac{1}{2z} ∑_{n=0}^∞(-1)^n (\frac{1}{z})^n- \frac{1}{6} ∑_{n=0}^∞(-1)^n (\frac{z}{3})^n$$
d) $$\frac{1}{2} ∑_{n=0}^∞(-1)^n (\frac{1}{z})^n- \frac{1}{4} ∑_{n=0}^∞(-1)^n (\frac{z}{3})^n$$

Explanation:
Given: $$f(z)= \frac{1}{(z+1)(z+3)} = \frac{A}{z+1}+ \frac{B}{z+3}$$
1=A(z+3)+ B (z+1)
Solving by Partial Fractions,
$$f(z)= \frac{1}{2}(\frac{1}{z+1}))- \frac{1}{2}(\frac{1}{z+3})…………… (2)$$
Given region is 1 < |z| < 3
$$|\frac{1}{z}|$$ <1, $$|\frac{z}{3}|$$ <1
From (2),
$$f(z)= \frac{1}{2} \frac{1}{[1+\frac{1}{z}]} -\frac{1}{2(3)} \frac{1}{[1+\frac{z}{3}]}$$
$$= \frac{1}{2z}[1+\frac{1}{z}]^{-1}- \frac{1}{6} [1+\frac{z}{3}]^{-1}$$
$$f(z) = \frac{1}{2z} ∑_{n=0}^∞(-1)^n (\frac{1}{z})^n- \frac{1}{6} ∑_{n=0}^∞(-1)^n (\frac{z}{3})^n$$

8. Which of the following is obtained by evaluating $$\frac{z}{(z+1)(z+2)}$$ in Laurent’s Series valid for the region |z|<1?
a) $$(-1)∑_{n=0}^∞(-1)^n z^n+ ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n$$
b) $$∑_{n=0}^∞z^n- ∑_{n=0}^∞(\frac{z}{2})^n$$
c) $$∑_{n=0}^∞(\frac{1}{z})^n- ∑_{n=0}^∞(\frac{z}{2})^n$$
d) $$\frac{1}{z-1}-2∑_{n=0}^∞(z-1)^n$$

Explanation:
$$Let f(z)= \frac{z}{(z+1)(z+2)}$$
f(z) is not analytic at z=-1 and z=-2
$$\frac{z}{(z+1)(z+2)} = \frac{A}{z+1}+ \frac{B}{z+2}$$
z=A(z+2)+ B(z+1)
Solving by Partial Fractions,
$$f(z)= \frac{z}{(z+1)(z+2)} = \frac{-1)}{z+1}+ \frac{2}{z+2}$$
Given region is |z|<1
$$f(z)= \frac{-1}{1+z}+ \frac{2}{2(1+\frac{z}{2})}$$ [|z|< 1 also $$|\frac{z}{2}|$$<1].
$$=(-1)(1+z)^{-1}+ (1+ \frac{z}{2})^{-1}$$
$$f(z)=(-1)∑_{n=0}^∞(-1)^n z^n+ ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n$$

9. Which of the following is obtained by evaluating $$\frac{z}{z^2-3z+2}$$ valid for the region |z-1| < 1 using Laurent’s Series?
a) $$(-1)∑_{n=0}^∞(-1)^n z^n+ ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n$$
b) $$∑_{n=0}^∞z^n- ∑_{n=0}^∞(\frac{z}{2})^n$$
c) $$∑_{n=0}^∞(\frac{1}{z}^n- ∑_{n=0}^∞(\frac{z}{2})^n$$
d) $$\frac{1}{z-1}-2∑_{n=0}^∞(z-1)^n$$

Explanation:
$$f(z)= \frac{z}{z^2-3z+2}= \frac{z}{(z-1)(z-2)}= \frac{A}{z-1}+ \frac{B}{z-2}$$
z=A(z-2)+ B(z-1)
Solving by Partial Fractions,
$$f(z)= \frac{-1}{z-1}+ \frac{2}{z-2}= \frac{1}{1-z}+ \frac{2}{z-2}$$
Given region is | z-1 | < 1
Let u= z-1
z = u + 1, |u|<1
$$f(z)= \frac{-1}{u}+ \frac{2}{u-1}= \frac{-1}{u}- \frac{2}{1-u}= \frac{-1}{u}-2(1-u)^{-1}= \frac{-1}{u}-2∑_{n=0}^∞u^n$$
$$f(z)= \frac{1}{z-1}-2∑_{n=0}^∞(z-1)^n$$

10. Which of the following is obtained by evaluating $$f(z)= \frac{7z-2}{z(z-2)(z+1)}$$ valid in 1<|z+1|<3 using Laurent’s Series?
a) $$(-1)∑_{n=0}^∞(-1)^n z^n+ ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n$$
b) $$∑_{n=0}^∞z^n- ∑_{n=0}^∞(\frac{z}{2})^n$$
c) $$∑_{n=0}^∞(\frac{1}{z})^n- ∑_{n=0}^∞(\frac{z}{2})^n$$
d) $$\frac{-2}{z+1}+∑_{n=0}^∞\frac{1}{(z+1)^n} – \frac{2}{3} ∑_{n=0}^∞\frac{(z+1)^n}{3^n}$$

Explanation:
$$f(z)= \frac{7z-2)}{z(z-2)(z+1)}$$
$$\frac{7z-2}{z(z-2)(z+1)}= \frac{A}{z}+ \frac{B}{z-2}+ \frac{C}{z+1}$$
7z-2 = A(z-2)(z+1)+ Bz(z+1)+ Cz(z-2)
Solving by Partial Fractions,
$$f(z)= \frac{1}{z}+ \frac{2}{z-2}-\frac{3}{z+1}$$
Given region is 1 < |z+1|< 3
Let u = z + 1
$$|\frac{1}{u}|$$<1 and $$|\frac{u}{3}|<1$$
$$f(z)= \frac{1}{u-1}+ \frac{2}{u-3}- \frac{3}{u}= \frac{1}{u[1-\frac{1}{u}]} + \frac{2}{-3[1-\frac{u}{3}]}- \frac{3}{u}$$
$$f(z)= \frac{1}{u}[1-\frac{1}{u}]^{-1}- \frac{2}{3}[1-\frac{u}{3}]^{-1}- \frac{3}{u}$$
$$f(z)= \frac{1}{u} ∑_{n=0}^∞(\frac{1}{u})^n- \frac{2}{3} ∑_{n=0}^∞(\frac{u}{3})^n-(\frac{3}{u})$$
$$f(z)= \frac{-2}{z+1}+ ∑_{n=0}^∞\frac{1}{(z+1)^n} – \frac{2}{3} ∑_{n=0}^∞\frac{(z+1)^n}{3^n}$$

11. Which of the following is obtained by evaluating $$\frac{1}{(z-1)(z-2)}$$ valid for the region |z-1|<1 using Laurent’s Series?
a) $$(-1)∑_{n=0}^∞(-1)^n z^n+ ∑_{n=0}^∞(-1)^n(\frac{z}{2})^n$$
b) $$∑_{n=0}^∞z^n- ∑_{n=0}^∞(\frac{z}{2})^n$$
c) $$\frac{-2}{z+1}+ ∑_{n=0}^∞\frac{1}{(z+1)^n} -\frac{2}{3} ∑_{n=0}^∞\frac{(z+1)^n}{3^n}$$
d) $$\frac{-1}{z-1}- ∑_{n=0}^∞(z-1)^n$$

Explanation:
Given: $$f(z)= \frac{1}{(z-1)(z-2)}= \frac{A}{z-1}+ \frac{B}{z-2}$$
1=A(z-2)+B(z-1)
Solving by Partial Fractions,
$$f(z)= \frac{1}{(z-1)(z-2)}= \frac{-1}{z-1}+ \frac{1}{z-2}$$
Given region is |z-1|< 1
Let u = z-1; z = u + 1
|u|<1
$$f(z)= \frac{-1}{u}+ \frac{1}{u-1}= \frac{-1}{u}- \frac{1}{1-u}= \frac{-1}{u}-(1-u)^{-1}$$
$$= \frac{-1}{u}- ∑_{n=0}^∞u^n$$
$$f(z)= \frac{-1}{z-1}- ∑_{n=0}^∞(z-1)^n$$

12. Which of the following is obtained by evaluating $$\frac{z^2-1}{z^2+5z+6}$$ valid in the region |z| >3 using Laurent’s Series?
a) $$1+(\frac{3}{z}) ∑_{n=0}^∞(-1)^n (\frac{2}{z})^n- \frac{8}{z} ∑_{n=0}^∞(-1)^n (\frac{3}{2})^n$$
b) $$(-1)∑_{n=0}^∞(-1)^n z^n+ ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n$$
c) $$\frac{-2}{z+1}+ ∑_{n=0}^∞\frac{1}{(z+1)^n}- \frac{2}{3} ∑_{n=0}^∞\frac{(z+1)^n}{3^n}$$
d) $$∑_{n=0}^∞z^n-∑_{n=0}^∞(\frac{z}{2})^n$$

Explanation:
$$f(z)= \frac{z^2-1}{z^2+ 5z+6}= \frac{z^2-1}{(z+2)(z+3)}$$
$$\frac{z^2-1}{z^2+ 5z+6}= 1- \frac{5z+7}{(z+2)(z+3)} ………….. (1)$$
$$\frac{5z+7}{(z+2)(z+3)}= \frac{A}{z+2}+ \frac{B}{z+3} ………………. (2)$$
5z+7=A(z+3)+B(z+2)
Solving by Partial fractions, we get
$$\frac{5z+7}{(z+2)(z+3)}= \frac{-3}{z+2}+ \frac{8}{z+3}$$
Substituting in (1)
$$f(z)=\frac{z^2-1}{z^2+ 5z+6}=1+ \frac{3}{z+2}- \frac{8}{z+3} …………………… (3)$$
Given region is | z | > 3
$$|\frac{3}{z}|$$<1; $$|\frac{2}{z}|$$< 1
$$f(z)= 1+ \frac{3}{z}[\frac{1}{1+\frac{2}{z}}]- \frac{8}{z}[\frac{1}{1+\frac{3}{z}}]$$
$$=1+ \frac{3}{z}[1+ \frac{2}{z}]^{-1}- \frac{8}{z} [1+\frac{3}{z}]^{-1}$$
$$f(z)= 1+(\frac{3}{z}) ∑_{n=0}^∞(-1)^n (\frac{2}{z})^n- \frac{8}{z} ∑_{n=0}^∞(-1)^n (\frac{3}{2})^n$$

13. Which of the following is obtained by evaluating $$\frac{z}{z^2-3z+2}$$ in the region 1< |z| <2 using Laurent’s Series?
a) $$(-1)∑_{n=0}^∞(-1)^n z^n+ ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n$$
b) $$(-1)∑_{n=0}^∞z^n+ ∑_{n=0}^∞(-1)^n$$
c) $$\frac{-2}{z+1}+∑_{n=0}^∞\frac{1}{(z+1)^n}- \frac{2}{3} ∑_{n=0}^∞\frac{(z+1)^n}{3^n}$$
d) $$\frac{-1}{z} ∑_{n=0}^∞(\frac{1}{z})^n- ∑_{n=0}^∞(\frac{z}{2})^n$$

Explanation:
$$f(z)= \frac{z}{z^2-3z+2}= \frac{z}{(z-1)(z-2)}= \frac{A}{z-1}+ \frac{B}{z-2}$$
z=A(z-2)+ B(z-1)
Solving by Partial Fractions,
$$f(z)= \frac{-1}{z-1}+ \frac{2}{z-2}= \frac{1}{1-z}+ \frac{2}{z-2}$$
Given region is 1 < |z| < 2
1<|z| →$$|\frac{1}{z}|$$<1
|z|<2→$$|\frac{z}{2}|$$<1
$$f(z)= \frac{-1}{z}[\frac{1}{1-\frac{1}{z}}]+ \frac{1}{(-2)} \frac{2}{(1-\frac{z}{2})}$$
$$= \frac{-1}{z}[1-\frac{1}{z}]^{-1}- [1-\frac{z}{2}]^{-1}$$
$$f(z)= \frac{-1}{z}∑_{n=0}^∞(\frac{1}{z})^n- ∑_{n=0}^∞(\frac{z}{2})^n$$

14. Which of the following is obtained by evaluating $$\frac{z}{(z+1)(z+2)}$$ in the region 1 < |z| < 2 using Laurent’s Series?
a) $$1+(\frac{3}{z}) ∑_{n=0}^∞(-1)^n (\frac{2}{z})^n- \frac{8}{z} ∑_{n=0}^∞(-1)^n (\frac{3}{2})^n$$
b) $$(-1)∑_{n=0}^∞z^n+ ∑_{n=0}^∞(-1)^n$$
c) $$\frac{-1}{z} ∑_{n=0}^∞(\frac{1}{z})^n- ∑_{n=0}^∞(\frac{z}{2})^n$$
d) $$\frac{-1}{z} ∑_{n=0}^∞(-1)^n (\frac{1}{z})^n+ ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n$$

Explanation: Let $$f(z)= \frac{z}{(z+1)(z+2)}$$
f(z) is not analytic at z=-1 and z=-2
$$\frac{z}{(z+1)(z+2)} = \frac{A}{z+1}+ \frac{B}{z+2}$$
z=A(z+2)+ B(z+1)
Solving by Partial Fractions,
$$f(z)=\frac{z}{(z+1)(z+2)} = \frac{-1}{z+1}+ \frac{2}{z+2}$$
Given region is 1 < |z| < 2
$$|\frac{1}{z}|$$<1 and $$|\frac{z}{2}|$$<1
$$f(z)= \frac{-1}{z} ∑_{n=0}^∞(-1)^n (\frac{1}{z})^n+ ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n$$

15. Which of the following is obtained by evaluating $$\frac{1}{(z-1)(z-2)}$$ valid for the region 1<|z|< 2 using Laurent’s Series?
a) $$(-1)∑_{n=0}^∞z^n+ ∑_{n=0}^∞(-1)^n$$
b) $$\frac{-2}{z+1}+∑_{n=0}^∞\frac{1}{z+1)^n}- \frac{2}{3} ∑_{n=0}^∞\frac{(z+1)^n}{3^n}$$
c) $$∑_{n=0}^∞z^n- ∑_{n=0}^∞(\frac{z}{2})^n$$
d) $$\frac{-1}{z} ∑_{n=0}^∞(\frac{1}{z})^n- \frac{1}{2} ∑_{n=0}^∞(\frac{z}{2})^n$$

Explanation:
Given: $$f(z)= \frac{1}{(z-1)(z-2)}= \frac{A}{z-1}+ \frac{B}{z-2}$$
1 = A(z-2) + B(z-1)
Solving by Partial Fractions,
$$f(z)= \frac{1}{(z-1)(z-2)}= \frac{-1}{z-1}+ \frac{1}{z-2}$$
Given region is 1 < |z| < 2
|1/z|< 1 and |z/2|<1
$$f(z)= \frac{-1}{z[1-\frac{z}{2}]} + \frac{1}{-2(1-\frac{z}{2})} = \frac{-1}{z}[1-\frac{1}{z}]^{-1}- \frac{1}{2}[1-\frac{z}{2}]^{-1}$$
$$f(z)= \frac{-1}{z} ∑_{n=0}^∞(\frac{1}{z})^n- \frac{1}{2} ∑_{n=0}^∞(\frac{z}{2})^n$$

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