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64-1-6-10-1-10-I-need-so-much-plz-




Question Number 197299 by bbbbbbbb last updated on 12/Sep/23
((−64))^(1/6) −((−10))^(1/(10)) =?  I need so much plz
$$\sqrt[{\mathrm{6}}]{−\mathrm{64}}−\sqrt[{\mathrm{10}}]{−\mathrm{10}}=? \\ $$$$\boldsymbol{{I}}\:\boldsymbol{\mathrm{need}}\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{much}}\:\boldsymbol{\mathrm{plz}} \\ $$
Answered by Frix last updated on 13/Sep/23
∀n∈N\{0}: −n=ne^(iπ)  ⇒  ((−n))^(1/k) =(n)^(1/k) e^(i(π/k)) =(n)^(1/k) (cos (π/k) +i sin (π/k))
$$\forall{n}\in\mathbb{N}\backslash\left\{\mathrm{0}\right\}:\:−{n}={n}\mathrm{e}^{\mathrm{i}\pi} \:\Rightarrow \\ $$$$\sqrt[{{k}}]{−{n}}=\sqrt[{{k}}]{{n}}\mathrm{e}^{\mathrm{i}\frac{\pi}{{k}}} =\sqrt[{{k}}]{{n}}\left(\mathrm{cos}\:\frac{\pi}{{k}}\:+\mathrm{i}\:\mathrm{sin}\:\frac{\pi}{{k}}\right) \\ $$

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