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Question Number 132633 by mohammad17 last updated on 15/Feb/21

f i n d a l l v a l u e o f z i t s s a y t i s f i e s

\frac{1}{4 z - z^{3}} = \frac{1}{4 z} + \underset{n = 0}{\sum^{\infty}} \frac{z^{n}}{4^{n + 2}}

Answered by TheSupreme last updated on 15/Feb/21

\underset{n = 0}{\sum^{\infty}} \frac{z^{n}}{4^{n + 2}} = \frac{1}{16} \underset{n = 0}{\sum^{\infty}} {(\frac{z}{4})}^{n} = \frac{1}{16} (\frac{1}{1 - \frac{z}{4}}) = \frac{1}{16} (\frac{4}{4 - z})

\frac{1}{4 z - z^{3}} = \frac{1}{4 z} + \frac{1}{4 (4 - z)}

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

\frac{4 (4 - z) - (4 - z) (4 - z^{2}) - z (4 - z^{2})}{z (2 - z) (2 + z) 4 (4 - z)} = 0

16 - 4 z - (16 - 4 z^{2} - 4 z + z^{3}) - 4 z + z^{3} = 0

4 z^{2} - 4 z = 0

4 z (z - 1) = 0

z = 1

z = 0 \dots o u t o f D o m a i n

z = 1