已知复数Z=COSa+iSINa1.Z的平方,Z的立方,Z的四次方2.猜想Z的N次方并用数学归纳法证明
2019-04-14
已知复数Z=COSa+iSINa
1.Z的平方,Z的立方,Z的四次方
2.猜想Z的N次方并用数学归纳法证明
优质解答
z = cosx + isinx
z² = (cosx + isinx)²
= cos²x + i²sin²x + 2isinxcosx
= cos²x - sin²x + 2isinxcosx
= cos2x + isin2x
z³ = (cosx + isinx)³ = (cosx + isinx)²(cosx + isinx)
= (cos2x + isin2x)(cosx + isinx)
= cos2xcosx + i²sin2xsinx + isin2xcosx + icos2xsinx
= cos2xcosx - sin2xsinx + i(sin2xcosx + cos2xsinx)
= cos3x + isin3x
z⁴ = (cosx + isinx)⁴ = (cosx + isinx)³(cosx + isinx)
= (cos3x + isin3x)(cosx + isinx)
= cos3xcosx + i²sin3xsinx + isin3xcosx + icos3xsinx
= cos3xcosx - sin3xsinx + i(sin3xcosx + cos3xsinx)
= cos4x + isin4x
设命题P(n)：zⁿ = (cosx + isinx)ⁿ = cos(nx) + isin(nx)
当n = 1时明显成立,P(1)成立
假设当对于某些正整数k,P(k)都成立,
i.e. z^k = cos(kx) + isin(kx)
当n = k + 1时,LHS = z^(k + 1) = z^k · z
= [cos(kx) + isin(kx)][cosx + isinx]
= cos(kx)cosx + i²sin(kx)sinx + isin(kx)cosx + icos(kx)sinx
= cos(kx)cosx - sin(kx)sinx + i[sin(kx)cosx + cos(kx)sinx]
= cos[(k + 1)x] + isin[(k + 1)x]
= RHS
P(k + 1)亦成立.
所以根据数学归纳法原理,对于任何正整数n,P(n)都成立.
这就是De Moivre's Theorem
z = cosx + isinx
z² = (cosx + isinx)²
= cos²x + i²sin²x + 2isinxcosx
= cos²x - sin²x + 2isinxcosx
= cos2x + isin2x
z³ = (cosx + isinx)³ = (cosx + isinx)²(cosx + isinx)
= (cos2x + isin2x)(cosx + isinx)
= cos2xcosx + i²sin2xsinx + isin2xcosx + icos2xsinx
= cos2xcosx - sin2xsinx + i(sin2xcosx + cos2xsinx)
= cos3x + isin3x
z⁴ = (cosx + isinx)⁴ = (cosx + isinx)³(cosx + isinx)
= (cos3x + isin3x)(cosx + isinx)
= cos3xcosx + i²sin3xsinx + isin3xcosx + icos3xsinx
= cos3xcosx - sin3xsinx + i(sin3xcosx + cos3xsinx)
= cos4x + isin4x
设命题P(n)：zⁿ = (cosx + isinx)ⁿ = cos(nx) + isin(nx)
当n = 1时明显成立,P(1)成立
假设当对于某些正整数k,P(k)都成立,
i.e. z^k = cos(kx) + isin(kx)
当n = k + 1时,LHS = z^(k + 1) = z^k · z
= [cos(kx) + isin(kx)][cosx + isinx]
= cos(kx)cosx + i²sin(kx)sinx + isin(kx)cosx + icos(kx)sinx
= cos(kx)cosx - sin(kx)sinx + i[sin(kx)cosx + cos(kx)sinx]
= cos[(k + 1)x] + isin[(k + 1)x]
= RHS
P(k + 1)亦成立.
所以根据数学归纳法原理,对于任何正整数n,P(n)都成立.
这就是De Moivre's Theorem