Question Number 71326 by sadimuhmud 136 last updated on 13/Oct/19
$$\left(−\mathrm{64}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} =?\left(\boldsymbol{\mathrm{I}}\mathrm{s}\:\mathrm{there}\:\mathrm{any}\:\mathrm{short}\:\mathrm{cut}\:\mathrm{for}\:\mathrm{mcq}\right) \\ $$
Answered by MJS last updated on 13/Oct/19
$$\mathrm{2}^{\mathrm{6}} =\mathrm{64}\:\Rightarrow\:\sqrt[{\mathrm{6}}]{−\mathrm{64}}=\mathrm{2i} \\ $$
Commented by MJS last updated on 14/Oct/19
![this depends on what you have already learned. usually first we learn ((−r))^(1/3) =−(r)^(1/3) ∀r∈R^+ generally ((−r))^(1/(2n+1)) =−(r)^(1/(2n+1)) ∀n∈Z ∀r∈R^+ which indeed is the only exception of the rule (z)^(1/n) =((re^(iθ) ))^(1/n) =(r)^(1/n) e^(i(θ/n)) in C without this rule ((−64))^(1/6) =(((−64))^(1/3) )^(1/2) =((−4))^(1/2) =2i [not ((−64))^(1/6) =(((−64))^(1/2) )^(1/3) =((8i))^(1/3) which can′t be solved on this stage] within this rule ((−64))^(1/6) =((64e^(iπ) ))^(1/6) =((64))^(1/6) e^(i(π/6)) =2e^(i(π/6)) =(√3)+i](https://www.tinkutara.com/question/Q71365.png)
$$\mathrm{this}\:\mathrm{depends}\:\mathrm{on}\:\mathrm{what}\:\mathrm{you}\:\mathrm{have}\:\mathrm{already} \\ $$$$\mathrm{learned}. \\ $$$$\mathrm{usually}\:\mathrm{first}\:\mathrm{we}\:\mathrm{learn}\:\sqrt[{\mathrm{3}}]{−{r}}=−\sqrt[{\mathrm{3}}]{{r}}\:\forall{r}\in\mathbb{R}^{+} \\ $$$$\mathrm{generally}\:\sqrt[{\mathrm{2}{n}+\mathrm{1}}]{−{r}}=−\sqrt[{\mathrm{2}{n}+\mathrm{1}}]{{r}}\:\forall{n}\in\mathbb{Z}\:\forall{r}\in\mathbb{R}^{+} \\ $$$$\mathrm{which}\:\mathrm{indeed}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{exception}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{rule}\:\sqrt[{{n}}]{{z}}=\sqrt[{{n}}]{{r}\mathrm{e}^{\mathrm{i}\theta} }=\sqrt[{{n}}]{{r}}\mathrm{e}^{\mathrm{i}\frac{\theta}{{n}}} \:\mathrm{in}\:\mathbb{C} \\ $$$$ \\ $$$$\mathrm{without}\:\mathrm{this}\:\mathrm{rule} \\ $$$$\sqrt[{\mathrm{6}}]{−\mathrm{64}}=\sqrt[{\mathrm{2}}]{\sqrt[{\mathrm{3}}]{−\mathrm{64}}}=\sqrt[{\mathrm{2}}]{−\mathrm{4}}=\mathrm{2i} \\ $$$$\:\:\:\:\:\left[\mathrm{not}\:\sqrt[{\mathrm{6}}]{−\mathrm{64}}=\sqrt[{\mathrm{3}}]{\sqrt[{\mathrm{2}}]{−\mathrm{64}}}=\sqrt[{\mathrm{3}}]{\mathrm{8i}}\:\mathrm{which}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\right. \\ $$$$\left.\:\:\:\:\:\:\:\mathrm{solved}\:\mathrm{on}\:\mathrm{this}\:\mathrm{stage}\right] \\ $$$$ \\ $$$$\mathrm{within}\:\mathrm{this}\:\mathrm{rule} \\ $$$$\sqrt[{\mathrm{6}}]{−\mathrm{64}}=\sqrt[{\mathrm{6}}]{\mathrm{64e}^{\mathrm{i}\pi} }=\sqrt[{\mathrm{6}}]{\mathrm{64}}\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{6}}} =\mathrm{2e}^{\mathrm{i}\frac{\pi}{\mathrm{6}}} =\sqrt{\mathrm{3}}+\mathrm{i} \\ $$
Answered by Kunal12588 last updated on 14/Oct/19
$$\left(−\mathrm{1}\right)^{\mathrm{1}/\mathrm{6}} ×\left(\mathrm{2}^{\mathrm{6}} \right)^{\mathrm{1}/\mathrm{6}} ={i}^{\mathrm{1}/\mathrm{3}} ×\mathrm{2}=? \\ $$
Commented by MJS last updated on 14/Oct/19
![i^(1/3) =(e^(i(π/2)) )^(1/3) =e^(i(π/6)) =((√3)/2)+(1/2)i and i^(1/3) ≠(1/i^3 ) [z^(1/3) =(1/z^3 ) ⇔ z^(1/3) z^3 =1 ⇔ z^(4/3) =1 ⇔ z=1]](https://www.tinkutara.com/question/Q71364.png)
$$\mathrm{i}^{\mathrm{1}/\mathrm{3}} =\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \right)^{\mathrm{1}/\mathrm{3}} =\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{6}}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{and}\:\mathrm{i}^{\mathrm{1}/\mathrm{3}} \neq\frac{\mathrm{1}}{\mathrm{i}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\left[\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{3}}} =\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{3}} }\:\Leftrightarrow\:{z}^{\frac{\mathrm{1}}{\mathrm{3}}} {z}^{\mathrm{3}} =\mathrm{1}\:\Leftrightarrow\:{z}^{\frac{\mathrm{4}}{\mathrm{3}}} =\mathrm{1}\:\Leftrightarrow\:{z}=\mathrm{1}\right] \\ $$
Commented by Kunal12588 last updated on 14/Oct/19
ahhh, why they are so complex. yes because they are complex numbers! thanks sir.