Complex Variables and Partial Differential Equations (3130005)

Question-3c

BE | Semester-3 Summer-2020 | 27-10-2020

Q3) (c)

Expand $f (z) = \frac{1}{(z + 2) (z + 4)}$ valid for the region $(i) |z| < 2, (ii) 2 < |z| < 4$ and $(i) |z| > 4 .$

Here,

f (z) = \frac{1}{(z + 2) (z + 4)}

\Rightarrow f (z) = \frac{1}{2} [\frac{(z + 4) - (z + 2)}{(z + 2) (z + 4)}]

\Rightarrow f (z) = \frac{1}{2} [\frac{(z + 4)}{(z + 2) (z + 4)} - \frac{(z + 2)}{(z + 2) (z + 4)}]

\Rightarrow f (z) = \frac{1}{2} [\frac{1}{(z + 2)} - \frac{1}{(z + 4)}]_____(1)

(i) |z| < 2

i . e . |\frac{z}{2}| < 1 & |\frac{z}{4}| < 1

(1) \Rightarrow f (z) = \frac{1}{2} [\frac{1}{(z + 2)} - \frac{1}{(z + 4)}]

\Rightarrow f (z) = \frac{1}{2} [\frac{1}{2 (1 + \frac{z}{2})} - \frac{1}{4 (1 + \frac{z}{4})}]

\Rightarrow f (z) = \frac{1}{4} (1 - \frac{z}{2} + \frac{z^{2}}{4} - \frac{z^{3}}{8} + . . .) - \frac{1}{8} (1 - \frac{z}{4} + \frac{z^{2}}{16} - \frac{z^{3}}{64} + . . .)

(ii) 2 < |z| < 4

i . e . 2 < |z| & |z| < 4 \Rightarrow |\frac{2}{z}| < 1 & |\frac{z}{4}| < 1

(1) \Rightarrow f (z) = \frac{1}{2} [\frac{1}{(z + 2)} - \frac{1}{(z + 4)}]

\Rightarrow f (z) = \frac{1}{2} [\frac{1}{z (1 + \frac{2}{z})} - \frac{1}{4 (1 + \frac{z}{4})}]

\Rightarrow f (z) = \frac{1}{2 z} (1 - \frac{2}{z} + \frac{4}{z^{2}} - \frac{8}{z^{3}} + . . .) - \frac{1}{8} (1 - \frac{z}{4} + \frac{z^{2}}{16} - \frac{z^{3}}{64} + . . .)

(iii) |z| > 4

i . e . |\frac{z}{4}| > 1 \Rightarrow |\frac{4}{z}| < 1 & |\frac{z}{4}| < 1

(1) \Rightarrow f (z) = \frac{1}{2} [\frac{1}{(z + 2)} - \frac{1}{(z + 4)}]

\Rightarrow f (z) = \frac{1}{2} [\frac{1}{z (1 + \frac{2}{z})} - \frac{1}{z (1 + \frac{4}{z})}]

\Rightarrow f (z) = \frac{1}{2 z} (1 - \frac{2}{z} + \frac{4}{z^{2}} - \frac{8}{z^{3}} + . . .) - \frac{1}{2 z} (1 - \frac{4}{z} + \frac{16}{z^{2}} - \frac{64}{z^{3}} + . . .)

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