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Question Number 120679 by kaivan.ahmadi last updated on 02/Nov/20

$$h$$

Commented by bobhans last updated on 01/Nov/20

$$\sqrt[3]{-8}$$$$(1.0+1.732051\mathrm{i})$$

Commented by bobhans last updated on 01/Nov/20

$$\mathrm{the}\mathrm{result}\mathrm{from}\mathrm{calculator}$$

Commented by mathmax by abdo last updated on 02/Nov/20

$$\mathrm{if}\mathrm{a}>0\mathrm{and}\mathrm{a}\ne 1\mathrm{we}\mathrm{have}{\mathrm{a}}^{\mathrm{x}}={\mathrm{e}}^{\mathrm{xln}\left(\mathrm{a}\right)}$$$$\mathrm{if}\mathrm{a}<\mathrm{o}\Rightarrow {\mathrm{a}}^{\mathrm{x}}={(-(-\mathrm{a}))}^{\mathrm{x}}={(-1)}^{\mathrm{x}}{(-\mathrm{a})}^{\mathrm{x}}={\left({\mathrm{e}}^{\mathrm{i}\pi }\right)}^{\mathrm{x}}{\mathrm{e}}^{\mathrm{xln}(-\mathrm{a})}$$$$={\mathrm{e}}^{\mathrm{i}\pi \mathrm{x}}{\mathrm{e}}^{\mathrm{xln}(-\mathrm{a})}=(\cos \pi \mathrm{x}+\mathrm{isin}\left(\pi \mathrm{x}\right){\mathrm{e}}^{\mathrm{xln}(-\mathrm{a})}(-\mathrm{a}>0)$$$$\mathrm{or}{\mathrm{a}}^{\mathrm{x}}={\mathrm{e}}^{\mathrm{xln}(-\mathrm{a})+\mathrm{i}\pi \mathrm{x}}$$

Answered by MJS_new last updated on 01/Nov/20

$$\mathrm{but}{(-8)}^{\frac{1}{3}}\mathit{i}\mathit{s}\mathrm{defined}.$$$$\mathrm{there}\mathrm{are}2\mathrm{definitions}$$$$\left(1\right)z\in \mathbb{C},r\in {\mathbb{R}}^{+}:z^{\frac{1}{n}}={\left(r{\mathrm{e}}^{\mathrm{i}\theta }\right)}^{\frac{1}{n}}=r^{\frac{1}{n}}{\mathrm{e}}^{\mathrm{i}\frac{\theta }{n}}$$$$\left(2\right)\mathrm{for}\theta =\pi \wedge n=2k+1\mathrm{we}\mathrm{have}z=r{\mathrm{e}}^{\mathrm{i}\pi }=-r$$$$\mathrm{and}\mathrm{define}{z}^{\frac{1}{2k+1}}={(-r)}^{\frac{1}{2k+1}}=-\left(r^{\frac{1}{2k+1}}\right)$$$$\mathrm{this}{2}^{\mathrm{nd}}\mathrm{definition}\mathrm{is}‘‘{\mathrm{older}}^{″}\mathrm{because}\mathrm{in}$$$$\mathrm{basic}\mathrm{mathematics}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{common}\mathrm{to}\mathrm{use}\mathrm{it}$$$$\Rightarrow {(-8)}^{\frac{1}{3}}=\{\begin{array}{l}1+\sqrt{3}\mathrm{i}\\ -2\end{array}$$

Commented by bobhans last updated on 01/Nov/20

$$\mathrm{but}\mathrm{sir}\mathrm{why}\mathrm{if}\mathrm{we}\mathrm{use}\mathrm{a}\mathrm{calculator}$$$$\mathrm{give}\mathrm{the}\mathrm{answer}1+i\sqrt{3}?$$

Commented by MJS_new last updated on 02/Nov/20

$$\mathrm{depends}\mathrm{on}\mathrm{the}\mathrm{calculator}.\mathrm{the}\mathrm{ones}\mathrm{for}$$$$\mathrm{high}\mathrm{schools}\mathrm{should}\mathrm{give}-2,\mathrm{the}\mathrm{ones}\mathrm{for}$$$$\mathrm{university}\mathrm{give}1+\sqrt{3}\mathrm{i}$$$$\mathrm{I}\mathrm{guess}\mathrm{using}\mathrm{the}\mathrm{exceptions}\sqrt[3]{-27}=-3;$$$$\sqrt[5]{-32}=-2etc.\mathrm{makes}\mathrm{it}\mathrm{impossible}\mathrm{to}$$$$\mathrm{get}\mathrm{a}\mathrm{result}\mathrm{for}\mathrm{calculations}\mathrm{like}{(-1)}^{\frac{2}{3}}$$$$\mathrm{and}\mathrm{the}\mathrm{exeptions}\mathrm{do}\mathrm{harm}\mathrm{to}\mathrm{functions}$$$$\mathrm{like}f\left(x\right):y={(-1)}^x\mathrm{for}x\in \mathbb{R}\wedge y\in \mathbb{C}$$$$\Rightarrow \mathrm{most}\mathrm{advanced}\mathrm{calculators}\mathrm{use}$$$$z^{p+q\mathrm{i}}={\left(r{\mathrm{e}}^{\mathrm{i}\theta }\right)}^{p+q\mathrm{i}}=r^{p+q\mathrm{i}}{\mathrm{e}}^{-q\theta +\mathrm{ip}\theta }=$$$$=\frac{r^p}{{\mathrm{e}}^{q\theta }}{\mathrm{e}}^{i(p\theta +q\ln r)}\mathrm{which}\mathrm{for}q=0\mathrm{gives}{r}^p{\mathrm{e}}^{\mathrm{i}p\theta }$$

Commented by bobhans last updated on 02/Nov/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by malwaan last updated on 02/Nov/20

$${(-8)}^{\frac{1}{3}}isdefined$$$$x={(-8)}^{\frac{1}{3}}$$$$\Rightarrow x^3=-8\Rightarrow x^3+8=0$$$$(x+2)(x^2-2x+4)=0$$$$\Rightarrow x=-2$$$$andx=1\pm \sqrt{3}i$$

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