Complex Variables and Partial Differential Equations (3130005)

Question-3c-OR

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

Q3) (c)

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

Here,

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

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

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

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

(i) |z| < 1

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

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

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

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

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

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

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

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

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

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

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