is (−1)^(m/n) =((((−1 )^m ))^(1/n) ) or =(((−1))^(1/n) )^m or both of them are fault and why ?


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Question Number 86230 by M±th+et£s last updated on 27/Mar/20

$i s {(- 1)}^{\frac{m}{n}} = (\sqrt[n]{{(- 1)}^{m}}) o r = {(\sqrt[n]{- 1})}^{m}$ $o r b o t h o f t h e m a r e f a u l t a n d w h y ?$
	Answered by MJS last updated on 27/Mar/20
	$(−1)^(m/n) =(e^(iπ) )^(m/n) =e^(iπ(m/n)) =cos ((πm)/n) +i sin ((πm)/n) (((−1)^m ))^(1/n) =(((e^(iπ) )^m ))^(1/n) =(e^(iπm) )^(1/n) = { (((1)^(1/n) =1; m=2k)),((((−1))^(1/n) =e^(i(π/n)) =cos (π/n) +i sin (π/n); m=2k+1)) :}; k∈Z (((−1))^(1/n) )^m =((e^(iπ) )^(1/n) )^m =(e^(i(π/n)) )^m =e^(i((πm)/n)) =cos ((πm)/n) +i sin ((πm)/n)$
	${(- 1)}^{\frac{m}{n}} = {(e^{i π})}^{\frac{m}{n}} = e^{i π \frac{m}{n}} = \cos \frac{π m}{n} + i \sin \frac{π m}{n}$ $\sqrt[n]{{(- 1)}^{m}} = \sqrt[n]{{(e^{i π})}^{m}} = \sqrt[n]{e^{i π m}} = {\begin{cases} \sqrt[n]{1} = 1; m = 2 k \\ \sqrt[n]{- 1} = e^{i \frac{π}{n}} = \cos \frac{π}{n} + i \sin \frac{π}{n}; m = 2 k + 1 \end{cases}; k \in Z$ ${(\sqrt[n]{- 1})}^{m} = {(\sqrt[n]{e^{i π}})}^{m} = {(e^{i \frac{π}{n}})}^{m} = e^{i \frac{π m}{n}} = \cos \frac{π m}{n} + i \sin \frac{π m}{n}$
	Commented by M±th+et£s last updated on 27/Mar/20

	$t h a n k y o u s i r$
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