This set of Complex Integration Multiple Choice Questions & Answers (MCQs) focuses on “Taylor’s Series”.
1. Which of the following expansions is obtained by expanding ez as a Taylor’s series about z=0?
a) \(1+ \frac{z}{1!}+ \frac{z^2}{2!}+ \frac{z^3}{3!}+ … \)
b) \(\frac{z}{1!}+ \frac{z^2}{2!}+ \frac{z^3}{3!}+ … \)
c) \(1-\frac{z}{1!}+ \frac{z^2}{2!}- \frac{z^3}{3!}+⋯ \)
d) \(\frac{z}{1!}- \frac{z^2}{2!}-\frac{z^3}{3!}-… \)
View Answer
Explanation:
| Function | f(z)=ez | f'(z)=ez | f”(z)=ez | f”‘(z)=ez | …. |
|---|---|---|---|---|---|
| Value at z=0 | f(0)=1 | f'(0)=1 | f”(0)=1 | f”‘(0)=1 | …. |
Taylor’s Series about z=0 is
\(f(z)= f(0)+ \frac{f'(0)}{1!} z+ \frac{f”(0)}{2!}z^2+ \frac{(f”'(0)}{3!}z^3+⋯\)
\(f(z)= 1+ \frac{z}{1!}+ \frac{z^2}{2!}+ \frac{z^3}{3!}+⋯ \)
2. Which of the following expansions is obtained by expanding f (z) = log (1+z) as a Taylor’s Series about z=0 if |z|<1?
a)\( 1+\frac{z}{1!}+ \frac{z^2}{2!}+ \frac{z^3}{3!}+ … \)
b)\(1-\frac{z}{1!}+ \frac{z^2}{2!}- \frac{z^3}{3!}+⋯ \)
c)\(z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+⋯\)
d)\(z+\frac{z^2}{2}+\frac{z^3}{3}+\frac{z^4}{4}+⋯\)
View Answer
Explanation:
| Function | f(z)=log(1+z) | \(f'(z)=\frac{1}{1+z}\) | \(f”(z)=\frac{-1}{(1+z)^2}\) | \(f'”(z)=\frac{2}{(1+z)^3}\) | ……. |
| Value at z=0 | f(0)=0 | f'(0)=1 | f”(0)=-1 | f'”(0)=2 | …… |
Taylor’s Series about z=0 is \(f(z)= f(0)+ \frac{f'(0)}{1!}z+ \frac{f”(0)}{2!} z^2+\frac{f”'(0)}{3!} z^3+⋯\)
\(f(z)= 0+ \frac{1}{1!}z+\frac{(-1)}{2!}z^2+ \frac{2}{3!}z^3+⋯\)
If |z| 1 \(f(z)= z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+⋯ \)
3. Which of the following expansions is obtained by expanding \(\frac{1}{z-3}\) at z=1 as a Taylor’s series?
a) \(\frac{-1}{2}-\frac{1}{4}(z-1)-\frac{1}{8}(z-1)^2-\frac{1}{16}(z-1)^3-…\)
b) \(1-(z-1)+(z-1)^2-(z-1)^3+⋯\)
c) \(1+(z-1)+(z-1)^2+(z-1)^3+⋯\)
d) \(1+(z-1)^2+ (z-1)^4+⋯\)
View Answer
Explanation:
| Function | Value at z=1 | \(f(z)=\frac{1}{z-3}\) | \(f(1)=\frac{1}{1-3}=\frac{-1}{2}\) |
|---|---|
| \(f'(z)=\frac{-1}{(z-3)^2}\) | \(f(1)=\frac{-1}{(1-3)^2}=\frac{-1}{4}\) |
| \(f”(z)=\frac{2}{(z-3)^3}\) | \(f(1)=\frac{2}{(1-3)^3}=\frac{=1}{4}\) | \(f'”(z)=\frac{-6}{(z-3)^4}\) | \(f'”(z)=\frac{-6}{(1-3)^4}=\frac{-3}{-8}\) |
Taylor’s series about z=a is
\(f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯\)
\(f(z)= \frac{-1}{2}+\frac{(\frac{-1}{4})}{1!}(z-1)+\frac{(\frac{-1}{4})}{2!}(z-1)^2+\frac{(\frac{-3}{8})}{3!}(z-1)^3+⋯\)
\(f(z)= \frac{-1}{2}-\frac{1}{4}(z-1)-\frac{1}{8}(z-1)^2-\frac{1}{16}(z-1)^3+⋯ \)
4. Which of the following expansions is obtained by expanding \(\frac{1}{z}\) about z=i as a Taylor’s Series?
a) \(–i+(z-i)+i(z-i)^2-(z-i)^3+⋯\)
b) \(–i-(z-i)-i(z-i)^2-(z-i)^3-…\)
c) \(i+(z-i)+i(z-i)^2+ (z-i)^3+⋯\)
d) \(i-(z-i)-i(z-i)^2-(z-i)^3+⋯\)
View Answer
Explanation:
Taylor’s series about z=a is
\(f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯\)
\(f(z)= f(i)+\frac{f'(i)}{1!}(z-i)+\frac{f^”(i)}{2!}(z-i)^2+ \frac{f”'(i)}{3!}(z-i)^3+⋯\)
\(f(z)=(-i)+\frac{1}{1!}(z-i)+\frac{2i}{2!}(z-i)^2+\frac{(-6)}{3!}(z-i)^3+⋯\)
\(f(z)= -i+(z-i)+i(z-i)^2-(z-i)^3+⋯\)
5. Which of the following expansions is obtained by expanding \(\frac{1}{z^2}\) as a Taylor’s Series about the point z=2?
a) \(\frac{1}{2}-\frac{1}{2}(z-2)+ \frac{1}{16}(z-2)^2+\frac{1}{8}(z-2)^3+⋯\)
b) \(\frac{1}{4}-\frac{1}{4}(z-2)+ \frac{3}{16}(z-2)^2-\frac{1}{8}(z-2)^3+⋯\)
c) \(-\frac{1}{2}- \frac{1}{2}(z-2)+\frac{1}{16}(z-2)^2-\frac{1}{8}(z-2)^3+⋯\)
d) \(\frac{1}{4}- \frac{1}{2}(z-2)+\frac{1}{16}(z-2)^2+\frac{1}{32}(z-2)^3+⋯\)
View Answer
Explanation:
| Function | \(f(z)=\frac{1}{z^2}\) | \(f'(z)=\frac{-2}{z^3}\) | \(f”(z)=\frac{6}{z^4}\) | \(f”‘(z)=\frac{-25}{z^5}\) | … |
|---|---|---|---|---|---|
| Value at z=2 | \(f(2)=\frac{1}{4}\) | \(f'(2)=\frac{-1}{4}\) | \(f”(2)=\frac{3}{8}\) | \(f'”(2)=\frac{-3}{4}\) | … |
Taylor’s series about z=a is
\(f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯\)
\(f(z)= f(2)+ \frac{f'(2)}{1!}(z-2)+\frac{f”(2)}{2!}(z-2)^2+\frac{f”'(2)}{3!}(z-2)^3+⋯\)
\(f(z)= \frac{1}{4}-\frac{1}{4}(z-2)+ \frac{3}{16}(z-2)^2-\frac{1}{8}(z-2)^3+⋯\)
6. Which of the following expansions is obtained by expanding cos z about \(z = \frac{π}{3}\) in Taylor’s Series?
a) \( 1-\frac{√3}{2}(z-\frac{π}{3})- \frac{1}{4}(z-\frac{π}{3})^2+ \frac{√3}{12}(z-\frac{π}{3})^3+⋯\)
b) \( \frac{1}{2}-\frac{√3}{2}(z-\frac{π}{3})- \frac{1}{4}(z-\frac{π}{3})^2+ \frac{√3}{12}(z-\frac{π}{3})^3+⋯\)
c) \( \frac{1}{3}-\frac{√3}{8} (z-\frac{π}{3})- \frac{1}{2}(z-\frac{π}{3})^2+ \frac{√3}{8}(z-\frac{π}{3})^3+⋯\)
d) \(1-\frac{√3}{8}(z-\frac{π}{3})- \frac{1}{4}(z-\frac{π}{3})^2+ \frac{√3}{8}(z-\frac{π}{3})^3+⋯\)
View Answer
Explanation:
| Function | Value at \(z=\frac{π}{3}\) |
|---|---|
| f(z)=cosz | \(f(\frac{π}{3})=cos\frac{π}{3}= \frac{1}{2}\) |
| f'(z)=-sinz | \(f'(\frac{π}{3})= -sin(\frac{π}{3})= -\frac{√3}{2}\) |
| f”(z)=-cos z | \(f”(\frac{π}{3})=-cos(\frac{π}{3})= \frac{-1}{2}\) |
| f”‘(z)=sinz | \(f”‘(\frac{π}{3})=sin(\frac{π}{3})= \frac{√3}{2}\) |
Taylor’s series about z=a is
\(f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯\)
\(f(z)= f(\frac{π}{3})+ \frac{f'(\frac{π}{3})}{1!}(z-\frac{π}{3})+\frac{f”(\frac{π}{3})}{2!}(z-\frac{π}{3})^2+\frac{f”’ (\frac{π}{3})}{3!}(z-\frac{π}{3})^3+⋯\)
\(f(z)= \frac{1}{2}+ \frac{(\frac{-√3}{2})}{1!}(z-\frac{π}{3})+ \frac{(\frac{-1}{2})}{2!} (z-\frac{π}{3})^2+\frac{(\frac{√3}{2})}{3!}(z-\frac{π}{3})^3+⋯\)
\(f(z)= \frac{1}{2}- \frac{√3}{2}(z-\frac{π}{3})- \frac{1}{4}(z-\frac{π}{3})^2+ \frac{√3}{12}(z-\frac{π}{3})^3+⋯\)
7. Which of the following expansions is obtained by expanding \(\frac{1}{z}\) about z=1 as a Taylor’s Series?
a) \(\frac{1}{2}-\frac{1}{2}(z-2)+ \frac{1}{16}(z-2)^2+\frac{1}{8}(z-2)^3+⋯\)
b) \(1-(z-1)+ (z-1)^2-(z-1)^3+⋯\)
c) \(1+(z-1)+ (z-1)^2+ (z-1)^3+⋯\)
d) \(\frac{1}{4}-\frac{1}{4}(z-2)+ \frac{3}{16}(z-2)^2-\frac{1}{8}(z-2)^3+⋯\)
View Answer
Explanation:
Taylor’s series about z=a is
\(f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯\)
\(f(z)= f(1)+ \frac{f'(1)}{1!}(z-1)+\frac{f”(1)}{2!}(z-1)^2+ \frac{f”’ (1)}{3!}(z-1)^3+⋯\)
\(f(z)= 1+ \frac{-1}{1!}(z-1)+ \frac{2}{2!}(z-1)^2+ \frac{-6}{3!}(z-1)^3+⋯\)
\(f(z)= 1-(z-1)+(z-1)^2-(z-1)^3+⋯ \)
8. A function f (z) analytic inside a circle C with centre at a, can be expanded in the series
f(z)= \(f(a)+ f'(a)(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”‘(a)}{3!}(z-a)^3+⋯+⋯+\frac{f^n (a)}{n!}(z-a)^n\)+⋯to ∞, which is convergent at every point inside C.
a) True
b) False
View Answer
Explanation: Let ϒ be a Circle having z in its interior and concentric with the circle C and lying inside C, so that f (z) is analytic inside and on ϒ.
Let ω be any point on ϒ. We have |z-a| < | ω-a| i.e. \(|\frac{z-a}{ω-a}|\)< 1 ………… (2)
By Cauchy’s Integral Formula,
\(f(z)= \frac{1}{2 π i}∫_ϒ\frac{f(ω)}{ω-z}dω= \frac{1}{2 π i}∫_ϒ\frac{f(ω)}{(ω-a)-(z-a)} dω \)
\(= \frac{1}{2πi}∫_ϒ\frac{f(ω)}{ω -z}dω= \frac{1}{2πi}∫_γ\frac{f(ω)}{(ω-a)-(z-a)}dω\)
\( = \frac{1}{2πi}∫_γ\frac{f(ω)}{ω-a}[1-\frac{z-a}{ω-a}]^{-1}dω \)
\(= \frac{1}{2πi}∫_γ\frac{f(ω)}{ω-a}[1+(\frac{z-a}{ω-a})+(\frac{z-a}{ω-a})^2+⋯+(\frac{z-a}{ω-a})^n+⋯to ∞]dω \)
\(= \frac{1}{2πi}∫_γ[∑_{n=0}^∞\frac{f(ω).(z-a)^n}{(ω-a)^{n+1}}]dω= ∑_{n=0}^∞(z-a)^n \frac{1}{2πi}∫_γ\frac{f(ω)}{(ω-a)^{n+1}}dω\)
\(=∑_{n=0}^∞(z-a)^n\frac{f^n(a)}{n!}= ∑_{n=0}^∞\frac{f^n(a)}{n!}(z-a)^n \)
Applying Summation,
f(z)= \(f(a)+ f'(a)(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”‘(a)}{3!}(z-a)^3+⋯+⋯+\frac{f^n(a)}{n!}(z-a)^n\)+⋯to ∞
9. Which of the following series is known as Maclaurin’s Series?
a) \(f(z)= f(0)+ zf'(0)+ \frac{z^2}{2!}f”(0)+ \frac{z^3}{3!}f”‘(0)+⋯+to ∞\)
b) \(f(z)= f(0)+ f'(0)+ f”(0)+ f”‘(0)+⋯+to ∞\)
c) \(f(z)= zf(0)+ z^2f'(0)+ z^3f”(0)+ …+to ∞ \)
d) \(f(z)= f(a)+ zf'(a)+ \frac{z^2}{2!} f”(a)+ \frac{z^3}{3!}f”'(a)+⋯+to ∞\)
View Answer
Explanation:
Taylor’s series about z=a is
\(f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯\)
Taking a=0, Taylor’s Series reduces to
\(f(z)= f(0)+ zf'(0)+ \frac{z^2}{2!}f”(0)+ \frac{z^3}{3!}f”’ (0)+⋯+to ∞ \)
This series is known a Maclaurin’s Series.
10. Which of the following expansions is obtained by expanding \(\frac{1}{z}\) about z=2?
a) \(\frac{1}{2}- \frac{1}{2}(z-2)+ \frac{1}{16}(z-2)^2+ \frac{1}{8}(z-2)^3+⋯\)
b) \(\frac{1}{4}- \frac{1}{4}(z-2)+ \frac{3}{16}(z-2)^2- \frac{1}{8}(z-2)^3+⋯\)
c) \(\frac{1}{4}- \frac{1}{4}(z-2)+ \frac{3}{16}(z-2)^2- \frac{1}{8}(z-2)^3+⋯\)
d) \(\frac{1}{2}- \frac{1}{4}(z-2)+ \frac{1}{8}(z-2)^2- \frac{1}{16}(z-2)^3+⋯\)
View Answer
Explanation:
| Function | Value at z=2 |
|---|---|
| \(f(z)=\frac{1}{z}\) | \(f(2)=\frac{1}{2}\) |
| \(f'(z)=\frac{-1}{z^2}\) | \(f'(2)=\frac{-1}{4}\) |
| \(f”(z)=\frac{2}{z^3}\) | \(f”(2)=\frac{1}{4}\) |
| \(f”‘(z)=\frac{-6}{z^4}\) | \(f'”(2)=\frac{-3}{8}\) |
\(f(z)= \frac{1}{2}+ \frac{(\frac{-1}{4})}{1!}(z-2)+ \frac{(\frac{1}{4})}{2!}(z-2)^2+ \frac{(\frac{-3}{8})}{3!}(z-2)^3+⋯\)
\(f(z)=\frac{1}{2}- \frac{1}{4}(z-2)+ \frac{1}{8}(z-2)^2- \frac{1}{16}(z-2)^3+⋯\)
11. Taylor’s series about z=0 is
f(z)= \(f(0)+ \frac{f'(0)}{1!}z+ \frac{f”(0)}{2!}z^2+ \frac{f”‘(0)}{3!}z^3+⋯+⋯+\frac{f^n(0)}{n!}z^n\)+⋯to ∞
a) True
b) False
View Answer
Explanation:
A function f (z) analytic inside a circle C with centre at a, can be expanded in the series
\(f(z)= f(a)+ f'(a)(z-a)+ \frac{f”(a)}{2!}(z-a)^2+ \frac{f”‘(a)}{3!}(z-a)^3+⋯+⋯+\frac{f^n(a)}{n!}(z-a)^n+⋯to ∞ \), which is convergent at every point inside C.
Substituting a=0 in the above series,
\(f(z)= f(0)+ \frac{f'(0)}{1!}z+ \frac{f”(0)}{2!}z^2+ \frac{f”‘(0)}{3!}z^3+⋯+⋯+\frac{f^n(0)}{n!}z^n+⋯to ∞ \)
12. Which of the following expansions is obtained by expanding \(\frac{3}{z^2}\) about z=3 as a Taylor’s Series?
a) \(\frac{1}{4}-\frac{1}{4}(z-2)+ \frac{3}{16}(z-2)^2- \frac{1}{8}(z-2)^3+⋯\)
b) \(\frac{1}{3}- \frac{√3}{8}(z-\frac{π}{3})- \frac{1}{2}(z-\frac{π}{3})^2+ \frac{√3}{8}(z-\frac{π}{3})^3+⋯\)
c) \(1-\frac{1}{3}(z-3)+ \frac{1}{9}(z-3)^2- \frac{1}{27}(z-3)^3+⋯\)
d) \( \frac{1}{2}- \frac{1}{4}(z-2)+ \frac{1}{8}(z-2)^2- \frac{1}{16}(z-2)^3+⋯\)
View Answer
Explanation:
| Function | Value at z=3 |
|---|---|
| \(f(z)=\frac{3}{z}\) | \(f(3)=\frac{3}{3}=1\) |
| \(f'(z)=\frac{-3}{z^2}\) | \(f'(3)=\frac{-1}{3}\) |
| \(f”(z)=\frac{6}{z^2}\) | \(f”(3)=\frac{2}{9}\) |
| \(f”‘(z)=\frac{-18}{z^4}\) | \(f”‘(3)=\frac{-2}{9}\) |
Taylor’s series about z=a is
\(f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯\)
\(f(z)= f(3)+ \frac{f'(3)}{1!}(z-3)+ \frac{f”(3)}{2!}(z-3)^2+ \frac{f”‘(3)}{3!}(z-3)^3+⋯\)
\(f(z)= 1-\frac{1}{3}(z-3)+ \frac{1}{9}(z-3)^2- \frac{1}{27}(z-3)^3+⋯\)
13. Which of the following expansions is obtained by expanding \(\frac{e^z}{(z)}\) about z=2 as a Taylor’s Series?
a) \(1-\frac{1}{3}(z-3)+ \frac{1}{9}(z-3)^2- \frac{1}{27}(z-3)^3+⋯\)
b) \(1-(z-1)+ (z-1)^2-(z-1)^3+⋯\)
c) \(-\frac{1}{4}(z-1)- \frac{1}{8}(z-1)^2- \frac{1}{16}(z-1)^3-…\)
d) \(\frac{e^2}{2}- \frac{e^2}{4}(z-2)+ \frac{e^2}{16}(z-2)^2- \frac{e^2}{96}(z-2)^3+⋯\)
View Answer
Explanation:
| Function | Value at z=2 |
|---|---|
| \(f(z)=\frac{e^z}{z}\) | \(f(2)=\frac{e^2}{2}\) |
| \(f'(z)=\frac{-e^z}{z^2}\) | \(f'(2)=\frac{-e^2}{4}\) |
| \(f”(z)=\frac{e^z}{z^3}\) | \(f”(2)=\frac{e^2}{8}\) |
| \(f”‘(z)=\frac{-e^z}{z^4}\) | \(f”‘(2)=\frac{-e^2}{16}\) |
Taylor’s series about z=a is
\(f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯\)
\(f(z)= f(2)+ \frac{f'(2)}{1!}(z-2)+ \frac{f”(2)}{2!}(z-2)^2+ \frac{f”'(2)}{3!}(z-2)^3+⋯\)
\(f(z)= \frac{e^2}{2}- \frac{e^2}{4}(z-2)+ \frac{e^2}{16}(z-2)^2- \frac{e^2}{96}(z-2)^3+⋯\)
14. Which of the following expansions is obtained by expanding f(z) = sin z about \(z=\frac{π}{2}\) as a Taylor’s series?
a) \(1-\frac{√3}{2}(z-\frac{π}{3})- \frac{1}{4}(z-\frac{π}{3})^2+ \frac{√3}{12}(z-\frac{π}{3})^3+⋯\)
b) \(1+\frac{√3}{8}(z-\frac{π}{3})+ \frac{1}{4}(z-\frac{π}{3})^2+ \frac{√3}{8}(z-\frac{π}{3})^3+⋯\)
c) \(1-\frac{√3}{8}(z-\frac{π}{3})- \frac{1}{4}(z-\frac{π}{3})^2+ \frac{√3}{8}(z-\frac{π}{3})^3+⋯\)
d) \(1-\frac{1}{2}(z-\frac{π}{2})^2+⋯\)
View Answer
Explanation:
| Function | Value at\(z=\frac{π}{2}\) |
|---|---|
| f(z)=sinz | \(f(\frac{π}{2})=1\) |
| f'(z)=cosz | \(f'(\frac{π}{2})=0\) |
| f”(z)=-sinz | \(f”(\frac{π}{2})=-1\) |
| f'”(z)=-cosz | \(f”‘(\frac{π}{2})=0\) |
Taylor’s series about z=a is
\(f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯\)
\(f(z)= f(\frac{π}{2})+ \frac{f'(\frac{π}{2})}{1!}](z-\frac{π}{2})+ \frac{f”(\frac{π}{2})}{2!}(z-\frac{π}{2})^2+ \frac{f”‘(\frac{π}{2})}{3!}(z-\frac{π}{2})^3+⋯\)
\(f(z)= 1-\frac{1}{2}(z-\frac{π}{2})^2+⋯\)
15. Which of the following expansions is obtained by expanding \(\frac{1}{z-5}\) at z=3 as a Taylor’s Series?
a) \(\frac{z}{1!}+ \frac{z^2}{2!}+ \frac{z^3}{3!}+ …\)
b) \(1-\frac{1}{2}(z-\frac{π}{2})^2+⋯\)
c) \(1-(z-1)+ (z-1)^2-(z-1)^3+⋯\)
d) \(\frac{-1}{2}- \frac{1}{4}(z-3)- \frac{1}{4}(z-3)^2- \frac{1}{16}(z-3)^3+⋯\)
View Answer
Explanation:
| Function | Value at z=3 |
|---|---|
| \(f(z)=\frac{1}{z-5}\) | \(f(3)=\frac{-1}{2}\) |
| \(f'(z)=\frac{-1}{(z-5)^2}\) | \(‘f(3)=\frac{-1}{4}\) |
| \(f”(z)=\frac{2}{(z-5)^3}\) | \(f”(3)=\frac{-1}{4}\) |
| \(f”‘(z)=\frac{-6}{(z-5)^4}\) | \(f”‘(3)=\frac{-3}{8}\) |
Taylor’s series about z=a is
\(f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯\)
\(f(z)=f(3)+ \frac{f'(3)}{1!}(z-3)+ \frac{f”(3)}{2!}(z-3)^2+ \frac{f”‘(3)}{3!}(z-3)^3+⋯\)
\(f(z)= \frac{-1}{2}- \frac{1}{4}(z-3)- \frac{1}{4}(z-3)^2- \frac{1}{16}(z-3)^3+⋯\)
Sanfoundry Global Education & Learning Series – Complex Integration.
To practice all areas of Complex Integration, here is complete set of 1000+ Multiple Choice Questions and Answers.
- Practice Probability and Statistics MCQ
- Check Engineering Mathematics Books
- Apply for 1st Year Engineering Internship
- Practice Numerical Methods MCQ
