(1+z)^n =(1−z)^n where z is a complex number


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Question Number 139052 by EnterUsername last updated on 21/Apr/21

${(1 + z)}^{n} = {(1 - z)}^{n}$ $w h e r e z i s a c o m p l e x n u m b e r$
	Answered by mathmax by abdo last updated on 21/Apr/21

	$z = - 1 is not solution let z \neq - 1$ $e \Rightarrow {(\frac{1 - z}{1 + z})}^{n} = 1 = e^{i (2 k π)} \Rightarrow \frac{1 - z}{1 + z} = e^{i (\frac{2 k π}{n})} k \in [[0, n - 1]]$ $\Rightarrow 1 - z = e^{\frac{2 ik π}{n}} + e^{\frac{2 ik π}{n}} z \Rightarrow 1 - e^{\frac{2 ik π}{n}} = (1 + e^{\frac{2 ik π}{n}}) z \Rightarrow$ $z_{k} = \frac{1 - e^{\frac{2 ik π}{n}}}{1 + e^{\frac{2 ik π}{n}}} = \frac{1 - \cos (\frac{2 k π}{n}) - isin (\frac{2 k π}{n})}{1 + \cos (\frac{2 k π}{n}) + isin (\frac{2 k π}{n})}$ $= \frac{{2 \sin}^{2} (\frac{k π}{n}) - 2 isin (\frac{k π}{n}) \cos (\frac{k π}{n})}{{2 \cos}^{2} (\frac{k π}{n}) + 2 isin (\frac{k π}{n}) \cos (\frac{k π}{n})} = \frac{- isin (\frac{k π}{n}) e^{\frac{ik π}{n}}}{\cos (\frac{k π}{n}) e^{\frac{ik π}{n}}} = - itan (\frac{k π}{n})$ $so the solution of this equation are z_{k} = - itan (\frac{k π}{n})$ $k \in [0, n - 1]]$
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