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Question Number 80084 by behi83417@gmail.com last updated on 30/Jan/20

$$-1={(-1)}^1={(-1)}^{\frac{2}{2}}={\left({(-1)}^2\right)}^{\frac{1}{2}}={\left(1\right)}^{\frac{1}{2}}=$$$$=\sqrt{1}=1$$$$\mathrm{what}\mathrm{do}\mathrm{you}\mathrm{think}\mathrm{about}\mathrm{this}?$$

Commented by key of knowledge last updated on 30/Jan/20

$${(-1)}^{\frac{2}{2}}={\left({(-1)}^{\frac{1}{2}}\right)}^2\ne {\left({(-1)}^2\right)}^{\frac{1}{2}}$$

Commented by behi83417@gmail.com last updated on 31/Jan/20

$$\mathrm{thanks}\mathrm{alot}{\mathrm{sir}}^{\prime }\mathrm{s}.$$$$\mathrm{perfect}\mathrm{explanation}.$$

Commented by MJS last updated on 31/Jan/20

$${\mathrm{we}}^{\prime }\mathrm{ve}\mathrm{got}\mathrm{a}\mathrm{definition}:$$$$\mathbb{C}=\{z={\mathrm{e}}^{\mathrm{i}\theta }r\mid -\pi <\theta ⩽\pi ;r\in {\mathbb{R}}^{+}\}$$$$\mathrm{we}\mathrm{know}\mathrm{that}{\mathrm{e}}^{\mathrm{i}\theta }={\mathrm{e}}^{\mathrm{i}(\theta +2\pi n)}\mathrm{with}n\in \mathbb{Z}\mathrm{but}\mathrm{we}$$$$\mathrm{only}\mathrm{use}\mathrm{this}\mathrm{for}\mathrm{finding}\mathrm{all}\mathrm{possible}\mathrm{solutions}$$$$\mathrm{of}\mathrm{certain}\mathrm{equations}$$$$\mathrm{we}\mathrm{define}$$$$z^q={\left({\mathrm{e}}^{\mathrm{i}\theta }r\right)}^q={\mathrm{e}}^{\mathrm{i}(\pi -\mathrm{mod}(\pi -q\theta ,2\pi )}{r}^q\mathrm{for}q\in \mathbb{Q},\mathbb{R},\mathbb{C}$$$$\mathrm{with}\mathrm{mod}(x,y)⩾0\mathrm{i}.\mathrm{e}.\mathrm{mod}(-3,7)=4$$$$[{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{only}\mathrm{one}\mathrm{exception}\mathrm{for}z\in {\mathbb{R}}^{+}\mathrm{and}$$$$n\in \mathbb{Z}:{(-z)}^{\frac{1}{2n+1}}=-\left(z^{\frac{1}{2n+1}}\right)\mathrm{This}\mathrm{is}\mathrm{useful}\mathrm{in}$$$$\mathrm{elementary}\mathrm{mathematics},\mathrm{like}3\mathrm{d}\mathrm{geometry}$$$$\mathrm{or}\mathrm{solving}\mathrm{polynomes}\mathrm{of}{3}^{\mathrm{rd}}\mathrm{degree}\mathrm{using}$$$$\mathrm{Cardano}.]$$$$\Rightarrow {\left(z^q\right)}^{\frac{1}{q}}\ne {\left(z^{\frac{1}{q}}\right)}^q\mathrm{in}\mathrm{many}\mathrm{cases}$$$${\left({(-1)}^2\right)}^{\frac{1}{2}}={\left(1\right)}^{\frac{1}{2}}=1$$$${\left({(-1)}^{\frac{1}{2}}\right)}^2={\left(\mathrm{i}\right)}^2=-1$$$${\left({(3+4\mathrm{i})}^4\right)}^{\frac{1}{4}}=4-3\mathrm{i}$$$${\left({(3+4\mathrm{i})}^{\frac{1}{4}}\right)}^4=3+4\mathrm{i}$$

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