下面等式怎样变形得到老师们讲解下吧

下面等式怎样变形得到老师们讲解下吧
下面等式怎样变形得到

老师们讲解下吧

下面等式怎样变形得到老师们讲解下吧
(1+i)^n
=[√2(√2/2+√2/2i)]^n
=2^(n/2)(cosπ/4+isinπ/4)^n
=2^(n/2)(cosnπ/4+isinnπ/4)
(1-i)^n
=(1+i)^n(1-i)^n/(1+i)^n
=(1-i^2)^n/(1+i)^n
=2^n/[2^(n/2)(cosnπ/4+isinnπ/4)
=2^(n/2)(cosnπ/4-isinnπ/4)/[(cosnπ/4+isinnπ/4)(cosnπ/4-isinnπ/4)]
=2^(n/2)(cosnπ/4-isinnπ/4)/[(cosnπ/4)^2-i^2(sinnπ/4)^2]
=2^(n/2)(cosnπ/4-isinnπ/4)
an=-2^[(n+1)/2][cos(n+1)π/4+isin(n+1)π/4]-2^(n+1)[cos(n+1)π/4-isin(n+1)π/4]
=-2^[(n+1)/2][cos(n+1)π/4+isin(n+1)π/4+cos(n+1)π/4-isin(n+1)π/4]
=-2^[(n+1)/2]*2cos(n+1)π/4
=-2^[(n+3)/2]cos(n+1)π/4
证毕

将复数表示成三角形式 ：r(cosθ+i sinθ) (T1) //: r--复数的模；θ--复数的辐角
1+i=√2[cos(π/4)+i sin(π/4)] (1)
1-i=√2[cos(π/4) -i sin(π/4)] (2)
根据棣美弗定理： [r(cosθ+i sinθ)]^n = r^n(co...

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将复数表示成三角形式 ：r(cosθ+i sinθ) (T1) //: r--复数的模；θ--复数的辐角
1+i=√2[cos(π/4)+i sin(π/4)] (1)
1-i=√2[cos(π/4) -i sin(π/4)] (2)
根据棣美弗定理： [r(cosθ+i sinθ)]^n = r^n(cosnθ+i sinnθ) （T2）
(1+i)^(n+1)=(√2)^(n+1)[cos{(n+1)π/4}+i sin{(n+1)π/4}] (3)
(1-i)^(n+1)=(√2)^(n+1)[cos{(n+1)π/4} - i sin{(n+1)π/4}] (4)
将(3)、(4)相加，对结果再取负号，得到：
an=-[(1+i)^(n+1)+(1-i)^(n+1)]=
=-2×(√2)^(n+1) cos[(n+1)π/4]
=-2×2^[(n+1)/2] cos[(n+1)π/4]
an = -2^[(n+3)/2] cos[(n+1)π/4] (5)

此即欲求结果。

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这道题考察复数基本运算和棣莫弗公式的运用。可作如下解答:
(1 i)^n=[√2(√2/2 √2/2i)]^n
=2^(n/2)(cosπ/4 isinπ/4)^n
=2^(n/2)(cosnπ/4 isinnπ/4)
(1-i)^n=(1 i)^n(1-i)^n/(1 i)^n
=(1-i^2)^n/(1 i)^n
=2^n/[2^(n/2)(co...

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这道题考察复数基本运算和棣莫弗公式的运用。可作如下解答:
(1 i)^n=[√2(√2/2 √2/2i)]^n
=2^(n/2)(cosπ/4 isinπ/4)^n
=2^(n/2)(cosnπ/4 isinnπ/4)
(1-i)^n=(1 i)^n(1-i)^n/(1 i)^n
=(1-i^2)^n/(1 i)^n
=2^n/[2^(n/2)(cosnπ/4 isinnπ/4)
=2^(n/2)(cosnπ/4-isinnπ/4)/[(cosnπ/4 isinnπ/4)(cosnπ/4-isinnπ/4)]
=2^(n/2)(cosnπ/4-isinnπ/4)/[(cosnπ/4)^2-i^2(sinnπ/4)^2]
=2^(n/2)(cosnπ/4-isinnπ/4)
an=-2^[(n 1)/2][cos(n 1)π/4 isin(n 1)π/4]-2^(n 1)[cos(n 1)π/4-isin(n 1)π/4]
=-2^[(n 1)/2][cos(n 1)π/4 isin(n 1)π/4 cos(n 1)π/4-isin(n 1)π/4]
=-2^[(n 1)/2]*2cos(n 1)π/4
=-2^[(n 3)/2]cos(n 1)π/4
只是运算较繁。

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