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) \( 1+(\frac{3}{2}) ∑_{n=0}^∞(-1)^n (\frac{2}{3})^n-\frac{8}{z} ∑_{n=0}^∞(-1)^n (\frac{3}{z})^n \)
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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) \( -\frac{3}{2} ∑_{n=0}^∞(-1)^n (\frac{2}{3})^n+ \frac{4}{5} ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n \)
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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) \( \frac{1}{2} ∑_{n=0}^∞(-1)^n (\frac{1}{z})^n-\frac{1}{4} ∑_{n=0}^∞(-1)^n (\frac{z}{3})^n \)
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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
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Answer: a
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) \( \frac{1}{z-1} – 2∑_{n=0}^∞(z-1)^n \)
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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) \( 1+(\frac{3}{2}) ∑_{n=0}^∞(-1)^n (\frac{2}{3})^n- \frac{8}{z} ∑_{n=0}^∞(-1)^n (\frac{3}{z})^n \)
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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) \( \frac{1}{2} ∑_{n=0}^∞(-1)^n (\frac{1}{z})^n- \frac{1}{4} ∑_{n=0}^∞(-1)^n (\frac{z}{3})^n \)
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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) \( \frac{1}{z-1}-2∑_{n=0}^∞(z-1)^n \)
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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) \( \frac{1}{z-1}-2∑_{n=0}^∞(z-1)^n \)
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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) \( \frac{-2}{z+1}+∑_{n=0}^∞\frac{1}{(z+1)^n} – \frac{2}{3} ∑_{n=0}^∞\frac{(z+1)^n}{3^n} \)
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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) \(\frac{-1}{z-1}- ∑_{n=0}^∞(z-1)^n \)
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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) \(∑_{n=0}^∞z^n-∑_{n=0}^∞(\frac{z}{2})^n \)
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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) \(\frac{-1}{z} ∑_{n=0}^∞(\frac{1}{z})^n- ∑_{n=0}^∞(\frac{z}{2})^n \)
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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) \( \frac{-1}{z} ∑_{n=0}^∞(-1)^n (\frac{1}{z})^n+ ∑_{n=0}^∞(-1)^n (\frac{z}{2})^n \)
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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) \( \frac{-1}{z} ∑_{n=0}^∞(\frac{1}{z})^n- \frac{1}{2} ∑_{n=0}^∞(\frac{z}{2})^n\)
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