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

((−1+i(√3)))^(1/6)  =....??

1+i36=....??

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] .

the6emerootsof1+i3arethecomplexzkwichverifyzk6=1+i3letsolvez6=1+i3z=reiθ1+i3∣=21+i3=2(12+i32)=2ei2π3soz6=1+i3r6ei(6θ)=2ei2π3r=216and6θ=2π3+2kπr=62θ=π9+kπ3k[[0,5]]zk=62ei(π9+kπ3)withk[0,5].

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)

1+3i=2e2π3i(reφi)q=rqeqφi(2e2π3i)16=216eπ9i=216cosπ9+216isinπ9

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