z^4=8sqrt(2)+8sqrt(2)i

解答

z^{4} = 82 + 82 i

z = 2 + 2 + 2 + i 2 - 2 + 2, z = - 2 - 2 + 2 + i 2 + 2 + 2, z = - 2 + 2 + 2 - i 2 - 2 + 2, z = 2 - 2 + 2 - i 2 + 2 + 2

求解步骤

z^{4} = 82 + 82 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 = 4, a = 82 + 82 i

∣ a ∣ = 16

ar g (a) = \frac{π}{4}

z = 416 (cos (\frac{\frac{π}{4} + 2 \cdot 0 π}{4}) + i sin (\frac{\frac{π}{4} + 2 \cdot 0 π}{4})), z = 416 (cos (\frac{\frac{π}{4} + 2 \cdot 1 π}{4}) + i sin (\frac{\frac{π}{4} + 2 \cdot 1 π}{4})), z = 416 (cos (\frac{\frac{π}{4} + 2 \cdot 2 π}{4}) + i sin (\frac{\frac{π}{4} + 2 \cdot 2 π}{4})), z = 416 (cos (\frac{\frac{π}{4} + 2 \cdot 3 π}{4}) + i sin (\frac{\frac{π}{4} + 2 \cdot 3 π}{4}))

化简

416 (cos (\frac{\frac{π}{4} + 2 \cdot 0 π}{4}) + i sin (\frac{\frac{π}{4} + 2 \cdot 0 π}{4})) : 2 + 2 + 2 + i 2 - 2 + 2

化简

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

化简

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

化简

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

z = 2 + 2 + 2 + i 2 - 2 + 2, z = - 2 - 2 + 2 + i 2 + 2 + 2, z = - 2 + 2 + 2 - i 2 - 2 + 2, z = 2 - 2 + 2 - i 2 + 2 + 2

x^{3} - 7 x^{2} + 4 x + 30 = 0

(x) = x^{3} - 3 x^{2} - 9 x - 5

so l v e f or x, x^{2} + (y - c b r t (x^{2}))^{2} = 25

roo t s z^{4} - 3 z^{3} - 5 z^{2} + 6 z - 8

3 x^{3} - 24 x^{2} + 36 x = 0