solve (1+ix)^n =n with x unknown real and n integr natural .


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Question Number 47543 by maxmathsup by imad last updated on 11/Nov/18

$s o l v e {(1 + i x)}^{n} = n w i t h x u n k n o w n r e a l a n d n i n t e g r n a t u r a l .$
	Answered by arcana last updated on 11/Nov/18

	$l n (e^{(1 + i x) n}) = n l n (e e^{i x}) = n$ $l n (e) + l n (e^{i x}) = 1 + l n (e^{i x}) = 1$ $l n (e^{i x}) = 0 = l n (1) \Rightarrow e^{i x} = c o s (x) + i s i n (x) = 1 + i 0$ $si y solo si c o s (x) = 1, s i n (x) = 0$ $x = k π con k \in Z$ $verificar deduccion$
	Commented by Smail last updated on 11/Nov/18

	${(1 + i x)}^{n} = l n (e^{{(1 + i x)}^{n}}) n o t l n (e^{(1 + i x) n})$
	Answered by Smail last updated on 11/Nov/18

	${(1 + i x)}^{n} = {n e}^{2 i k π} w i t h k = (0, 1, 2, . . ., n - 1)$ $1 + i x = \sqrt[n]{n} e^{\frac{2 i k π}{n}}$ $1 + i x = \sqrt[n]{n} c o s (\frac{2 k π}{n}) + i \sqrt[n]{n} s i n (\frac{2 k π}{n})$ $\sqrt[n]{n} c o s (\frac{2 k π}{n}) = 1 a n d \sqrt[n]{n} s i n (\frac{2 k π}{n}) = x$ $\frac{2 k π}{n} = {c o s}^{- 1} (\frac{1}{\sqrt[n]{n}}) + 2 m π w i t h m \in I N$ $x = \sqrt[n]{n} s i n ({c o s}^{- 1} (\frac{1}{\sqrt[n]{n}}) + 2 m π)$ $= \sqrt[n]{n} \sqrt{1 - \frac{1}{\sqrt[n]{n^{2}}}} = \sqrt{\sqrt[n]{n^{2}} - 1}$ $x = \sqrt{\sqrt[n]{n^{2}} - 1}$
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