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Question Number 35384 by 7991 last updated on 18/May/18

((−1+i(√3)))^(1/6)  =....??

$$\sqrt[{\mathrm{6}}]{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}\:=....?? \\ $$

Commented by prof Abdo imad last updated on 18/May/18

the 6^(eme)  roots of −1+i(√3)  are the complex z_k   wich verify z_k ^6 = −1+i(√3)   let solve z^6 =−1+i(√3)  z=r e^(iθ)       ∣−1+i(√3)∣=2 ⇒−1+i(√3) =2(−(1/2) +i((√3)/2))  =2e^(i((2π)/3))    so  z^6 =−1+i(√3) ⇔r^6  e^(i(6θ)) = 2 e^(i((2π)/3))  ⇒  r = 2^(1/6)    and 6θ= ((2π)/3) +2kπ ⇒r=^6 (√2)  θ= (π/9) +((kπ)/3)    k∈[[0,5]]⇒ z_k =^6 (√2)  e^(i((π/9) +((kπ)/3)))  with  k∈[0,5] .

$${the}\:\mathrm{6}^{{eme}} \:{roots}\:{of}\:−\mathrm{1}+{i}\sqrt{\mathrm{3}}\:\:{are}\:{the}\:{complex}\:{z}_{{k}} \\ $$$${wich}\:{verify}\:{z}_{{k}} ^{\mathrm{6}} =\:−\mathrm{1}+{i}\sqrt{\mathrm{3}}\:\:\:{let}\:{solve}\:{z}^{\mathrm{6}} =−\mathrm{1}+{i}\sqrt{\mathrm{3}} \\ $$$${z}={r}\:{e}^{{i}\theta} \:\:\:\:\:\:\mid−\mathrm{1}+{i}\sqrt{\mathrm{3}}\mid=\mathrm{2}\:\Rightarrow−\mathrm{1}+{i}\sqrt{\mathrm{3}}\:=\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\:\:{so}\:\:{z}^{\mathrm{6}} =−\mathrm{1}+{i}\sqrt{\mathrm{3}}\:\Leftrightarrow{r}^{\mathrm{6}} \:{e}^{{i}\left(\mathrm{6}\theta\right)} =\:\mathrm{2}\:{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\Rightarrow \\ $$$${r}\:=\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} \:\:\:{and}\:\mathrm{6}\theta=\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:+\mathrm{2}{k}\pi\:\Rightarrow{r}=^{\mathrm{6}} \sqrt{\mathrm{2}} \\ $$$$\theta=\:\frac{\pi}{\mathrm{9}}\:+\frac{{k}\pi}{\mathrm{3}}\:\:\:\:{k}\in\left[\left[\mathrm{0},\mathrm{5}\right]\right]\Rightarrow\:{z}_{{k}} =^{\mathrm{6}} \sqrt{\mathrm{2}}\:\:{e}^{{i}\left(\frac{\pi}{\mathrm{9}}\:+\frac{{k}\pi}{\mathrm{3}}\right)} \:{with} \\ $$$${k}\in\left[\mathrm{0},\mathrm{5}\right]\:. \\ $$

Answered by MJS last updated on 18/May/18

−1+(√3)i=2e^(((2π)/3)i)   (re^(ϕi) )^q =r^q e^(qϕi)   (2e^(((2π)/3)i) )^(1/6) =2^(1/6) e^((π/9)i) =2^(1/6) cos (π/9)+2^(1/6) i sin (π/9)

$$−\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}=\mathrm{2e}^{\frac{\mathrm{2}\pi}{\mathrm{3}}\mathrm{i}} \\ $$$$\left({r}\mathrm{e}^{\varphi\mathrm{i}} \right)^{{q}} ={r}^{{q}} \mathrm{e}^{{q}\varphi\mathrm{i}} \\ $$$$\left(\mathrm{2e}^{\frac{\mathrm{2}\pi}{\mathrm{3}}\mathrm{i}} \right)^{\frac{\mathrm{1}}{\mathrm{6}}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} \mathrm{e}^{\frac{\pi}{\mathrm{9}}\mathrm{i}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} \mathrm{cos}\:\frac{\pi}{\mathrm{9}}+\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} \mathrm{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{9}} \\ $$

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