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Question Number 35384 by 7991 last updated on 18/May/18
−1+i36=....??
Commented by prof Abdo imad last updated on 18/May/18
the6emerootsof−1+i3arethecomplexzkwichverifyzk6=−1+i3letsolvez6=−1+i3z=reiθ∣−1+i3∣=2⇒−1+i3=2(−12+i32)=2ei2π3soz6=−1+i3⇔r6ei(6θ)=2ei2π3⇒r=216and6θ=2π3+2kπ⇒r=62θ=π9+kπ3k∈[[0,5]]⇒zk=62ei(π9+kπ3)withk∈[0,5].
Answered by MJS last updated on 18/May/18
−1+3i=2e2π3i(reφi)q=rqeqφi(2e2π3i)16=216eπ9i=216cosπ9+216isinπ9
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