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Question Number 131344 by Algoritm last updated on 03/Feb/21

	Answered by bluberry508 last updated on 06/Feb/21

	$(1 + i) = \sqrt{2} e^{\frac{i π}{4} + 2 a i π}$ $(a \in Z)$ ${(1 + i)}^{n} = - \frac{1}{8} (1 + i)$ $(1 + i) [{(1 + i)}^{n - 1} + \frac{1}{8}] = 0$ $∴ {(1 + i)}^{n - 1} = - \frac{1}{8}$ $l e t n - 1 b e m$ ${(\sqrt{2} e^{\frac{i π}{4} + 2 a i π})}^{m} = (- 1) {(\frac{1}{2})}^{3} = e^{i π + 2 k i π} {(\frac{1}{2})}^{3}$ $(k \in Z)$ $2^{\frac{m}{2} + 3} e^{i π (\frac{m}{4} + 2 a m - 1 - 2 k)} = 1$ $∴ \frac{m}{2} + 3 = 0, \frac{m}{4} + 2 a m - 1 - 2 k = 2 p$ $(p \in Z)$ $m = n - 1 = - 6 t h u s, n = - 5$ $t h e n w e h a v e t h e f o l l o w i n g c o n d i t i o n$ $o f p a n d k .$ $2 p + 2 k + 12 a = - \frac{5}{2}$ $b u t t h i s c o n d i t i o n m u s t b e c o n t r a d i c t o r y$ $w i t h r e s p e c t t o p, k, a \in Z$ $t h e r e f o r e w e {c a n}^{'} t f i n d t h e v a l u e o f n$
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