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Question Number 86230 by M±th+et£s last updated on 27/Mar/20
is (−1)^(m/n) =((((−1 )^m ))^(1/n) ) or =(((−1))^(1/n) )^m or both of them are fault and why ?
$${is}\:\:\left(−\mathrm{1}\right)^{\frac{{m}}{{n}}} \:=\left(\sqrt[{{n}}]{\left(−\mathrm{1}\:\right)^{{m}} }\right)\:{or}\:=\left(\sqrt[{{n}}]{−\mathrm{1}}\right)^{{m}} \\ $$$${or}\:{both}\:{of}\:{them}\:{are}\:{fault}\:{and}\:{why}\:? \\ $$
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)
$$\left(−\mathrm{1}\right)^{\frac{{m}}{{n}}} =\left(\mathrm{e}^{\mathrm{i}\pi} \right)^{\frac{{m}}{{n}}} =\mathrm{e}^{\mathrm{i}\pi\frac{{m}}{{n}}} =\mathrm{cos}\:\frac{\pi{m}}{{n}}\:+\mathrm{i}\:\mathrm{sin}\:\frac{\pi{m}}{{n}} \\ $$$$\sqrt[{{n}}]{\left(−\mathrm{1}\right)^{{m}} }=\sqrt[{{n}}]{\left(\mathrm{e}^{\mathrm{i}\pi} \right)^{{m}} }=\sqrt[{{n}}]{\mathrm{e}^{\mathrm{i}\pi{m}} }=\begin{cases}{\sqrt[{{n}}]{\mathrm{1}}=\mathrm{1};\:{m}=\mathrm{2}{k}}\\{\sqrt[{{n}}]{−\mathrm{1}}=\mathrm{e}^{\mathrm{i}\frac{\pi}{{n}}} =\mathrm{cos}\:\frac{\pi}{{n}}\:+\mathrm{i}\:\mathrm{sin}\:\frac{\pi}{{n}};\:{m}=\mathrm{2}{k}+\mathrm{1}}\end{cases};\:{k}\in\mathbb{Z} \\ $$$$\left(\sqrt[{{n}}]{−\mathrm{1}}\right)^{{m}} =\left(\sqrt[{{n}}]{\mathrm{e}^{\mathrm{i}\pi} }\right)^{{m}} =\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{{n}}} \right)^{{m}} =\mathrm{e}^{\mathrm{i}\frac{\pi{m}}{{n}}} =\mathrm{cos}\:\frac{\pi{m}}{{n}}\:+\mathrm{i}\:\mathrm{sin}\:\frac{\pi{m}}{{n}} \\ $$
Commented by M±th+et£s last updated on 27/Mar/20
thank you sir
$${thank}\:{you}\:{sir}\: \\ $$

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