The Taylor series expansion of $\dfrac{1}{z-2}, |z|

Question

The Taylor series expansion of $\dfrac{1}{z-2}, |z|< 1$ is

TuteeHUB · Accepted Answer

$-\dfrac{1}{2}\left(1-\dfrac{z}{2} - \dfrac{z^2}{4}+ \dfrac{z^3}{8}...\right)$

TuteeHUB · Answer

$-\dfrac{1}{2}\left(1-\dfrac{z}{2} + \dfrac{z^2}{4}- \dfrac{z^3}{8}...\right)$

TuteeHUB · Answer

$-\dfrac{1}{2}\left(1+\dfrac{z}{2} + \dfrac{z^2}{4}+ \dfrac{z^3}{8}...\right)$

TuteeHUB · Answer

$x -\dfrac{1}{2}\left(1+\dfrac{z}{2} - \dfrac{z^2}{4}- \dfrac{z^3}{8}...\right)$

1.	The Taylor series expansion of \(\dfrac{1}{z-2}, \|z\|< 1\) is
A.	\(-\dfrac{1}{2}\left(1-\dfrac{z}{2} + \dfrac{z^2}{4}- \dfrac{z^3}{8}...\right)\)
B.	\(-\dfrac{1}{2}\left(1+\dfrac{z}{2} + \dfrac{z^2}{4}+ \dfrac{z^3}{8}...\right)\)
C.	\(-\dfrac{1}{2}\left(1-\dfrac{z}{2} - \dfrac{z^2}{4}+ \dfrac{z^3}{8}...\right)\)
D.	\(x -\dfrac{1}{2}\left(1+\dfrac{z}{2} - \dfrac{z^2}{4}- \dfrac{z^3}{8}...\right)\)
Answer» C. \(-\dfrac{1}{2}\left(1-\dfrac{z}{2} - \dfrac{z^2}{4}+ \dfrac{z^3}{8}...\right)\)

The Taylor series expansion of \(\dfrac{1}{z-2}, |z|< 1\) is