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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)}^{\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

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