# Complex Integration Questions and Answers – Taylor’s Series

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) $$\frac{z}{1!}- \frac{z^2}{2!}-\frac{z^3}{3!}-…$$

Explanation:

 Function Value at z=0 f(z)=ez f'(z)=ez f”(z)=ez f”‘(z)=ez …. f(0)=1 f'(0)=1 f”(0)=1 f”‘(0)=1 ….

$$f(z)= f(0)+ \frac{f'(0)}{1!} z+ \frac{f”(0)}{2!}z^2+ \frac{(f”'(0)}{3!}z^3+⋯$$
$$f(z)= 1+ \frac{z}{1!}+ \frac{z^2}{2!}+ \frac{z^3}{3!}+⋯$$

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)$$z+\frac{z^2}{2}+\frac{z^3}{3}+\frac{z^4}{4}+⋯$$

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) $$1+(z-1)^2+ (z-1)^4+⋯$$

Explanation:

Function Value at z=1
$$f(z)=\frac{1}{z-3}$$ $$f(1)=\frac{1}{1-3}=\frac{-1}{2}$$
$$f'(z)=\frac{-1}{(z-3)^2}$$ $$f(1)=\frac{-1}{(1-3)^2}=\frac{-1}{4}$$
$$f”(z)=\frac{2}{(z-3)^3}$$ $$f(1)=\frac{2}{(1-3)^3}=\frac{=1}{4}$$
$$f'”(z)=\frac{-6}{(z-3)^4}$$ $$f'”(z)=\frac{-6}{(1-3)^4}=\frac{-3}{-8}$$
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$$f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯$$
$$f(z)= \frac{-1}{2}+\frac{(\frac{-1}{4})}{1!}(z-1)+\frac{(\frac{-1}{4})}{2!}(z-1)^2+\frac{(\frac{-3}{8})}{3!}(z-1)^3+⋯$$ $$f(z)= \frac{-1}{2}-\frac{1}{4}(z-1)-\frac{1}{8}(z-1)^2-\frac{1}{16}(z-1)^3+⋯$$

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) $$i-(z-i)-i(z-i)^2-(z-i)^3+⋯$$

Explanation:
$$f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯$$
$$f(z)= f(i)+\frac{f'(i)}{1!}(z-i)+\frac{f^”(i)}{2!}(z-i)^2+ \frac{f”'(i)}{3!}(z-i)^3+⋯$$
$$f(z)=(-i)+\frac{1}{1!}(z-i)+\frac{2i}{2!}(z-i)^2+\frac{(-6)}{3!}(z-i)^3+⋯$$
$$f(z)= -i+(z-i)+i(z-i)^2-(z-i)^3+⋯$$

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) $$\frac{1}{4}- \frac{1}{2}(z-2)+\frac{1}{16}(z-2)^2+\frac{1}{32}(z-2)^3+⋯$$

Explanation:

 Function Value at z=2 $$f(z)=\frac{1}{z^2}$$ $$f'(z)=\frac{-2}{z^3}$$ $$f”(z)=\frac{6}{z^4}$$ $$f”‘(z)=\frac{-25}{z^5}$$ … $$f(2)=\frac{1}{4}$$ $$f'(2)=\frac{-1}{4}$$ $$f”(2)=\frac{3}{8}$$ $$f'”(2)=\frac{-3}{4}$$ …

$$f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯$$
$$f(z)= f(2)+ \frac{f'(2)}{1!}(z-2)+\frac{f”(2)}{2!}(z-2)^2+\frac{f”'(2)}{3!}(z-2)^3+⋯$$
$$f(z)= \frac{1}{4}-\frac{1}{4}(z-2)+ \frac{3}{16}(z-2)^2-\frac{1}{8}(z-2)^3+⋯$$

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) $$1-\frac{√3}{8}(z-\frac{π}{3})- \frac{1}{4}(z-\frac{π}{3})^2+ \frac{√3}{8}(z-\frac{π}{3})^3+⋯$$

Explanation:

Function Value at $$z=\frac{π}{3} f(z)=cosz [latex]f(\frac{π}{3})=cos⁡\frac{π}{3}= \frac{1}{2}$$
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 $$f”‘(\frac{π}{3})=sin⁡(\frac{π}{3})= \frac{√3}{2}$$

$$f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯$$
$$f(z)= f(\frac{π}{3})+ \frac{f'(\frac{π}{3})}{1!}(z-\frac{π}{3})+\frac{f”(\frac{π}{3})}{2!}(z-\frac{π}{3})^2+\frac{f”’ (\frac{π}{3})}{3!}(z-\frac{π}{3})^3+⋯$$
$$f(z)= \frac{1}{2}+ \frac{(\frac{-√3}{2})}{1!}(z-\frac{π}{3})+ \frac{(\frac{-1}{2})}{2!} (z-\frac{π}{3})^2+\frac{(\frac{√3}{2})}{3!}(z-\frac{π}{3})^3+⋯$$ $$f(z)= \frac{1}{2}- \frac{√3}{2}(z-\frac{π}{3})- \frac{1}{4}(z-\frac{π}{3})^2+ \frac{√3}{12}(z-\frac{π}{3})^3+⋯$$

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) $$\frac{1}{4}-\frac{1}{4}(z-2)+ \frac{3}{16}(z-2)^2-\frac{1}{8}(z-2)^3+⋯$$

Exlanation:
$$f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯$$
$$f(z)= f(1)+ \frac{f'(1)}{1!}(z-1)+\frac{f”(1)}{2!}(z-1)^2+ \frac{f”’ (1)}{3!}(z-1)^3+⋯$$
$$f(z)= 1+ \frac{-1}{1!}(z-1)+ \frac{2}{2!}(z-1)^2+ \frac{-6}{3!}(z-1)^3+⋯$$
$$f(z)= 1-(z-1)+(z-1)^2-(z-1)^3+⋯$$

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

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) $$f(z)= f(a)+ zf'(a)+ \frac{z^2}{2!} f”(a)+ \frac{z^3}{3!}f”'(a)+⋯+to ∞$$

Exlanation:
$$f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯$$
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) $$\frac{1}{2}- \frac{1}{4}(z-2)+ \frac{1}{8}(z-2)^2- \frac{1}{16}(z-2)^3+⋯$$

Explanation:

Function Value at z=2
$$f(z)=\frac{1}{z}$$ $$f(2)=\frac{1}{2}$$
$$f'(z)=\frac{-1}{z^2}$$ $$f'(2)=\frac{-1}{4}$$
$$f”(z)=\frac{2}{z^3}$$ $$f”(2)=\frac{1}{4}$$
$$f”‘(z)=\frac{-6}{z^4}$$ $$f'”(2)=\frac{-3}{8}$$
$$f(z)= f(2)+ \frac{f'(2)}{1!}(z-2)+ \frac{f”(2)}{2!}(z-2)^2+ \frac{f”‘(z)}{3!}(z-2)^3+⋯$$
$$f(z)= \frac{1}{2}+ \frac{(\frac{-1}{4})}{1!}(z-2)+ \frac{(\frac{1}{4})}{2!}(z-2)^2+ \frac{(\frac{-3}{8})}{3!}(z-2)^3+⋯$$
$$f(z)=\frac{1}{2}- \frac{1}{4}(z-2)+ \frac{1}{8}(z-2)^2- \frac{1}{16}(z-2)^3+⋯$$

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

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) $$\frac{1}{2}- \frac{1}{4}(z-2)+ \frac{1}{8}(z-2)^2- \frac{1}{16}(z-2)^3+⋯$$

Explanation:

Function Value at z=3
$$f(z)=\frac{3}{z}$$ $$f(3)=\frac{3}{3}=1$$
$$f'(z)=\frac{-3}{z^2}$$ $$f'(3)=\frac{-1}{3}$$
$$f”(z)=\frac{6}{z^2}$$ $$f”(3)=\frac{2}{9}$$
$$f”‘(z)=\frac{-18}{z^4}$$ $$f”‘(3)=\frac{-2}{9}$$

$$f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯$$
$$f(z)= f(3)+ \frac{f'(3)}{1!}(z-3)+ \frac{f”(3)}{2!}(z-3)^2+ \frac{f”‘(3)}{3!}(z-3)^3+⋯$$
$$f(z)= 1-\frac{1}{3}(z-3)+ \frac{1}{9}(z-3)^2- \frac{1}{27}(z-3)^3+⋯$$

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) $$\frac{e^2}{2}- \frac{e^2}{4}(z-2)+ \frac{e^2}{16}(z-2)^2- \frac{e^2}{96}(z-2)^3+⋯$$

Explanation:

Function Value at z=2
$$f(z)=\frac{e^z}{z}$$ $$f(2)=\frac{e^2}{2}$$
$$f'(z)=\frac{-e^z}{z^2}$$ $$f'(2)=\frac{-e^2}{4}$$
$$f”(z)=\frac{e^z}{z^3}$$ $$f”(2)=\frac{e^2}{8}$$
$$f”‘(z)=\frac{-e^z}{z^4}$$ $$f”‘(2)=\frac{-e^2}{16}$$

$$f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯$$
$$f(z)= f(2)+ \frac{f'(2)}{1!}(z-2)+ \frac{f”(2)}{2!}(z-2)^2+ \frac{f”'(2)}{3!}(z-2)^3+⋯$$
$$f(z)= \frac{e^2}{2}- \frac{e^2}{4}(z-2)+ \frac{e^2}{16}(z-2)^2- \frac{e^2}{96}(z-2)^3+⋯$$

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) $$1-\frac{1}{2}(z-\frac{π}{2})^2+⋯$$

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 $$f”‘(\frac{π}{2})=0$$

$$f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯$$
$$f(z)= f(\frac{π}{2})+ \frac{f'(\frac{π}{2})}{1!}](z-\frac{π}{2})+ \frac{f”(\frac{π}{2})}{2!}(z-\frac{π}{2})^2+ \frac{f”‘(\frac{π}{2})}{3!}(z-\frac{π}{2})^3+⋯$$ $$f(z)= 1-\frac{1}{2}(z-\frac{π}{2})^2+⋯$$

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) $$\frac{-1}{2}- \frac{1}{4}(z-3)- \frac{1}{4}(z-3)^2- \frac{1}{16}(z-3)^3+⋯$$

Explanation:

Function Value at z=3
$$f(z)=\frac{1}{z-5}$$ $$f(3)=\frac{-1}{2}$$
$$f'(z)=\frac{-1}{(z-5)^2}$$ $$‘f(3)=\frac{-1}{4}$$
$$f”(z)=\frac{2}{(z-5)^3}$$ $$f”(3)=\frac{-1}{4}$$
$$f”‘(z)=\frac{-6}{(z-5)^4}$$ $$f”‘(3)=\frac{-3}{8}$$

$$f(z)= f(a)+ \frac{f'(a)}{1!}(z-a)+ \frac{f”(a)}{2!}(z-a)^2+\frac{f”'(a)}{3!}(z-a)^3+⋯$$
$$f(z)=f(3)+ \frac{f'(3)}{1!}(z-3)+ \frac{f”(3)}{2!}(z-3)^2+ \frac{f”‘(3)}{3!}(z-3)^3+⋯$$
$$f(z)= \frac{-1}{2}- \frac{1}{4}(z-3)- \frac{1}{4}(z-3)^2- \frac{1}{16}(z-3)^3+⋯$$

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