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受欢迎的代数 >

z^3= i/8

解答

z^{3} = \frac{i}{8}

解答

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

求解步骤

z^{3} = \frac{i}{8}

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 = \frac{i}{8}

∣ a ∣ = \frac{1}{8}

ar g (a) = \frac{π}{2}

z = 3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{3})), z = 3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{3})), z = 3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}))

化简

3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{3})) : \frac{3}{4} + i \frac{1}{4}

化简

3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{3})) : - \frac{3}{4} + i \frac{1}{4}

化简

3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{3})) : - i \frac{1}{2}

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

流行的例子

x^{2} (x - 6) = 0

z^{4} = 1 + i

5 x^{3} - 6 x^{2} - 7 = 0

0 = x^{4} - 2 x^{2} - 8

solvefor y,x=y^6

so l v e f or y, x = y^{6}