I Cannot solve this taylor series problem of complex integration. Please help me sir **Transcription:**"Develop $ \frac{1}{c - z} $ in a Taylor series with center as $ z_0 = 2 $."

Given : f(z)=1c-z ,z0 =2 TAYLOR SERIES IS: f(z)=f(a)+f'(a)(z-a)+f''(a)(z-a)2/2 +..... f'(z)…

Answered: Develop 1 in a Taylor series with…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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I Cannot solve this taylor series problem of complex integration. Please help me sir

**Transcription:**

"Develop $ \frac{1}{c - z} $ in a Taylor series with center as $ z_0 = 2 $."

Expert Solution

Step 1

Given : $f (z) = \frac{1}{c - z}$ , $z_{0} = 2$

TAYLOR SERIES IS: $f (z) = f (a) + f' (a) (z - a) + f'' (a) {(z - a)}^{2} / 2 + . . . . .$

$f' (z) = \frac{- 1}{{(c - z)}^{2}} \times (- 1) = \frac{1}{{(c - z)}^{2}}$ $= \frac{1}{{(z - c)}^{2}}$ , $\Rightarrow f' (2) = \frac{1}{{(2 - c)}^{2}}$

$f (2) = \frac{1}{c - 2}$

$f " (z) = \frac{- 2}{{(z - c)}^{3}}$ $\Rightarrow f " (2) = \frac{- 2}{{(2 - c)}^{3}} = \frac{- 2}{{(- 1)}^{3} {(c - 2)}^{3}} = \frac{2}{{(c - 2)}^{3}}$

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