Complex Variables and Partial Differential Equations (3130005)

Question-2c

BE | Semester-3 Winter-2019 | 26-11-2019

Q2) (c)

Expand $f (z) = \frac{1}{(z - 1) (z + 2)}$ valid for the region $(I) |z| < 1$ , $(II) 1 < |z| < 2$ and $(III) |z| > 2$ .

Here,

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

\Rightarrow f (z) = \frac{1}{3} [\frac{(z + 2) - (z - 1)}{(z - 1) (z + 2)}]

\Rightarrow f (z) = \frac{1}{3} [\frac{(z + 2)}{(z - 1) (z + 2)} - \frac{(z - 1)}{(z - 1) (z + 2)}]

\Rightarrow f (z) = \frac{1}{3} [\frac{1}{(z - 1)} - \frac{1}{(z + 2)}] - - - - - (1)

(I) |z| < 1

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

(1) \Rightarrow f (z) = \frac{1}{3} [\frac{1}{(z - 1)} - \frac{1}{(z + 2)}]

\Rightarrow f (z) = \frac{1}{3} [- \frac{1}{(1 - z)} - \frac{1}{2 (\frac{z}{2} + 1)}]

\Rightarrow f (z) = - \frac{1}{3} [\frac{1}{(1 - z)} + \frac{1}{2 (1 + \frac{z}{2})}]

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

(II) 1 < |z| < 2

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

(1) \Rightarrow f (z) = \frac{1}{3} [\frac{1}{(z - 1)} - \frac{1}{(z + 2)}]

\Rightarrow f (z) = \frac{1}{3} [\frac{1}{z (1 - \frac{1}{z})} - \frac{1}{2 (\frac{z}{2} + 1)}]

\Rightarrow f (z) = \frac{1}{3} [\frac{1}{z (1 - \frac{1}{z})} - \frac{1}{2 (1 + \frac{z}{2})}]

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

\Rightarrow f (z) = \frac{1}{3} (\frac{1}{z} + \frac{1}{z^{2}} + \frac{1}{z^{3}} + \frac{1}{z^{4}} + . . .) - \frac{1}{6} (1 - \frac{z}{2} + \frac{z^{2}}{4} + \frac{z^{3}}{8} + . . .)

(III) |z| > 2

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

(1) \Rightarrow f (z) = \frac{1}{3} [\frac{1}{(z - 1)} - \frac{1}{(z + 2)}]

\Rightarrow f (z) = \frac{1}{3} [\frac{1}{z (1 - \frac{1}{z})} - \frac{1}{z (1 + \frac{2}{z})}]

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

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