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x*x*x*x+1=-1有解吗?
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x*x*x*x+1=-1有解吗?
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答案和解析
在实数域无解,复数域有四个解、且两两共轭.令:
Wk = r^(1/n) [cos (θ+2kπ)/n + i sin (θ+2kπ)/n] k = 0,1,2,...,n-1
本题中:r = 2,n = 4,θ = π,
x^4 = - 2+i 0 = 2(cos π + i sin π)
X0 = 2^(1/4) [cos (π/4) + i sin (π/4)] = 2^(1/4) √2/2 (1 + i )
X1 = 2^(1/4) [cos (3π/4) + i sin (3π/4)] = 2^(1/4) √2/2 (-1 + i)
X2 = 2^(1/4) [cos (5π/4) + i sin (5π/4)] = - 2^(1/4) √2/2 (1 + i)
X3 = 2^(1/4) [cos (7π/4) + i sin (7π/4)] = 2^(1/4) √2/2 (1 - i )
因此本题 x^4 = -2 的四个解为:X0,X1,X2,X3.
Wk = r^(1/n) [cos (θ+2kπ)/n + i sin (θ+2kπ)/n] k = 0,1,2,...,n-1
本题中:r = 2,n = 4,θ = π,
x^4 = - 2+i 0 = 2(cos π + i sin π)
X0 = 2^(1/4) [cos (π/4) + i sin (π/4)] = 2^(1/4) √2/2 (1 + i )
X1 = 2^(1/4) [cos (3π/4) + i sin (3π/4)] = 2^(1/4) √2/2 (-1 + i)
X2 = 2^(1/4) [cos (5π/4) + i sin (5π/4)] = - 2^(1/4) √2/2 (1 + i)
X3 = 2^(1/4) [cos (7π/4) + i sin (7π/4)] = 2^(1/4) √2/2 (1 - i )
因此本题 x^4 = -2 的四个解为:X0,X1,X2,X3.
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