作业帮下面等式怎样变形得到老师们讲解下吧
来源：学生作业帮 编辑：作业帮 分类：综合作业 时间：2024/05/19 23:16:18
下面等式怎样变形得到
老师们讲解下吧
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
再问： 2^(n/2)(cosπ/4 isinπ/4)^n =2^(n/2)(cosnπ/4 isinnπ/4)这里的n次方为什么拿到cos和sin中了是不是套公式在哪里找到的
再答： 这是复数转换成三角函数的基本公式啊, 你也可以用归纳法证明的 (cosπ/4+isinπ/4)^n = cosnπ/4+isinnπ/4 当n=1时, 显然上式成立 假设n=k, (cosπ/4+isinπ/4)^k = coskπ/4+isinkπ/4 则当n=k+1时 (cosπ/4+isinπ/4)^(k+1) =(cosπ/4+isinπ/4)^k*(cosπ/4+isinπ/4) =(coskπ/4+isinkπ/4)*(cosπ/4+isinπ/4) =coskπ/4cosπ/4+icoskπ/4sinπ/4+isinkπ/4cosπ/4+i^2sinkπ/4sinπ/4 =coskπ/4cosπ/4-sinkπ/4sinπ/4+i(sinkπ/4cosπ/4+coskπ/4sinπ/4) =cos(k+1)π/4+isin(k+1)π/4 由归纳法知, 假设成立 所以公式得证
=[√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
再问： 2^(n/2)(cosπ/4 isinπ/4)^n =2^(n/2)(cosnπ/4 isinnπ/4)这里的n次方为什么拿到cos和sin中了是不是套公式在哪里找到的
再答： 这是复数转换成三角函数的基本公式啊, 你也可以用归纳法证明的 (cosπ/4+isinπ/4)^n = cosnπ/4+isinnπ/4 当n=1时, 显然上式成立 假设n=k, (cosπ/4+isinπ/4)^k = coskπ/4+isinkπ/4 则当n=k+1时 (cosπ/4+isinπ/4)^(k+1) =(cosπ/4+isinπ/4)^k*(cosπ/4+isinπ/4) =(coskπ/4+isinkπ/4)*(cosπ/4+isinπ/4) =coskπ/4cosπ/4+icoskπ/4sinπ/4+isinkπ/4cosπ/4+i^2sinkπ/4sinπ/4 =coskπ/4cosπ/4-sinkπ/4sinπ/4+i(sinkπ/4cosπ/4+coskπ/4sinπ/4) =cos(k+1)π/4+isin(k+1)π/4 由归纳法知, 假设成立 所以公式得证