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Question Number 71326 by sadimuhmud 136 last updated on 13/Oct/19

$${(-64)}^{\frac{1}{6}}=?\left(\mathbf{I}\mathrm{s}\mathrm{there}\mathrm{any}\mathrm{short}\mathrm{cut}\mathrm{for}\mathrm{mcq}\right)$$

Answered by MJS last updated on 13/Oct/19

$$2^6=64\Rightarrow \sqrt[6]{-64}=2\mathrm{i}$$

Commented by MJS last updated on 14/Oct/19

$$\mathrm{this}\mathrm{depends}\mathrm{on}\mathrm{what}\mathrm{you}\mathrm{have}\mathrm{already}$$$$\mathrm{learned}.$$$$\mathrm{usually}\mathrm{first}\mathrm{we}\mathrm{learn}\sqrt[3]{-r}=-\sqrt[3]{r}\mathrm{\forall }r\in {\mathbb{R}}^{+}$$$$\mathrm{generally}\sqrt[2n+1]{-r}=-\sqrt[2n+1]{r}\mathrm{\forall }n\in \mathbb{Z}\mathrm{\forall }r\in {\mathbb{R}}^{+}$$$$\mathrm{which}\mathrm{indeed}\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{exception}\mathrm{of}\mathrm{the}$$$$\mathrm{rule}\sqrt[n]{z}=\sqrt[n]{r{\mathrm{e}}^{\mathrm{i}\theta }}=\sqrt[n]{r}{\mathrm{e}}^{\mathrm{i}\frac{\theta }{n}}\mathrm{in}\mathbb{C}$$$$$$$$\mathrm{without}\mathrm{this}\mathrm{rule}$$$$\sqrt[6]{-64}=\sqrt[2]{\sqrt[3]{-64}}=\sqrt[2]{-4}=2\mathrm{i}$$$$[\mathrm{not}\sqrt[6]{-64}=\sqrt[3]{\sqrt[2]{-64}}=\sqrt[3]{8\mathrm{i}}\mathrm{which}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{be}$$$$\mathrm{solved}\mathrm{on}\mathrm{this}\mathrm{stage}]$$$$$$$$\mathrm{within}\mathrm{this}\mathrm{rule}$$$$\sqrt[6]{-64}=\sqrt[6]{{64\mathrm{e}}^{\mathrm{i}\pi }}=\sqrt[6]{64}{\mathrm{e}}^{\mathrm{i}\frac{\pi }{6}}={2\mathrm{e}}^{\mathrm{i}\frac{\pi }{6}}=\sqrt{3}+\mathrm{i}$$

Answered by Kunal12588 last updated on 14/Oct/19

$${(-1)}^{1/6}\times {\left(2^6\right)}^{1/6}=i^{1/3}\times 2=?$$

Commented by MJS last updated on 14/Oct/19

$${\mathrm{i}}^{1/3}={\left({\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}\right)}^{1/3}={\mathrm{e}}^{\mathrm{i}\frac{\pi }{6}}=\frac{\sqrt{3}}{2}+\frac{1}{2}\mathrm{i}$$$$\mathrm{and}{\mathrm{i}}^{1/3}\ne \frac{1}{{\mathrm{i}}^3}$$$$[{\mathrm{z}}^{\frac{1}{3}}=\frac{1}{{\mathrm{z}}^3}\iff z^{\frac{1}{3}}{z}^3=1\iff z^{\frac{4}{3}}=1\iff z=1]$$

Commented by Kunal12588 last updated on 14/Oct/19

ahhh, why they are so complex. yes because they are complex numbers! thanks sir.

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