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受欢迎的代数 >

z^3=1-i

解答

z^{3} = 1 - i

解答

z = \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 - 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}}}{2} - i \frac{2 ^{\frac{2}{3}}}{2}

求解步骤

z^{3} = 1 - i

For

z^{n} = a

the solutions are

z_{k} = n ∣ a ∣ (cos (\frac{ar g ( a ) + 2 kπ}{n}) + i sin (\frac{ar g ( a ) + 2 kπ}{n})),

k = 0, 1, \dots, n - 1

对于

n = 3, a = 1 - i

∣ a ∣ = 2

ar g (a) = - \frac{π}{4}

z = 32 (cos (\frac{- \frac{π}{4} + 2 \cdot 0 π}{3}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 0 π}{3})), z = 32 (cos (\frac{- \frac{π}{4} + 2 \cdot 1 π}{3}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 1 π}{3})), z = 32 (cos (\frac{- \frac{π}{4} + 2 \cdot 2 π}{3}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 2 π}{3}))

化简

32 (cos (\frac{- \frac{π}{4} + 2 \cdot 0 π}{3}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 0 π}{3})) : \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 - 3 )}{4}

化简

32 (cos (\frac{- \frac{π}{4} + 2 \cdot 1 π}{3}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 1 π}{3})) : - \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}

化简

32 (cos (\frac{- \frac{π}{4} + 2 \cdot 2 π}{3}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 2 π}{3})) : - \frac{2 ^{\frac{2}{3}}}{2} - i \frac{2 ^{\frac{2}{3}}}{2}

z = \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 - 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}}}{2} - i \frac{2 ^{\frac{2}{3}}}{2}

流行的例子

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