((√3)−i)^(1+2i) =...


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Question Number 47389 by gunawan last updated on 09/Nov/18

${(\sqrt{3} - i)}^{1 + 2 i} = . . .$
Commented by maxmathsup by imad last updated on 09/Nov/18
$we have ∣(√3)−i∣=2 ⇒(√3)−i =2(((√3)/2)−(1/2))=2(cos(−(π/6))+isin(−(π/6))) =2e^(−((iπ)/6)) ⇒((√3)−i)^(1+2i) ={2e^(−((iπ)/6)) }^(1+2i) =2^(1+2i) e^(−((iπ(1+2i))/6)) =e^(−((iπ−2π)/6)) =e^(π/3) e^(−((iπ)/6)) 2^(1+2i) =e^(π/3) (((√3)/2) −(1/2))2(cos2 +isin2) = ((√3)−1)e^(π/3) (cos2 +isin2) .$
$w e h a v e ∣ \sqrt{3} - i ∣= 2 \Rightarrow \sqrt{3} - i = 2 (\frac{\sqrt{3}}{2} - \frac{1}{2}) = 2 (c o s (- \frac{π}{6}) + i s i n (- \frac{π}{6}))$ $= 2 e^{- \frac{i π}{6}} \Rightarrow {(\sqrt{3} - i)}^{1 + 2 i} = {2 e^{- \frac{i π}{6}}}^{1 + 2 i} = 2^{1 + 2 i} e^{- \frac{i π (1 + 2 i)}{6}}$ $= e^{- \frac{i π - 2 π}{6}} = e^{\frac{π}{3}} e^{- \frac{i π}{6}} 2^{1 + 2 i} = e^{\frac{π}{3}} (\frac{\sqrt{3}}{2} - \frac{1}{2}) 2 (c o s 2 + i s i n 2)$ $= (\sqrt{3} - 1) e^{\frac{π}{3}} (c o s 2 + i s i n 2) .$
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