复数 设a=(1+√3i)/2,b=(1-√3i)/2设a=(1+√3i)/2,b=(1-√3i)/2,1.求a²,a³,b²,b³的值 2.当n∈N*时,计算a^n+b^n第一问我会做a²=(-1+√3i)/2,a³=-1,b²=(-1-√3i)/2,b³=-1求解第二问,我的思路:a
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![复数 设a=(1+√3i)/2,b=(1-√3i)/2设a=(1+√3i)/2,b=(1-√3i)/2,1.求a²,a³,b²,b³的值 2.当n∈N*时,计算a^n+b^n第一问我会做a²=(-1+√3i)/2,a³=-1,b²=(-1-√3i)/2,b³=-1求解第二问,我的思路:a](/uploads/image/z/10267623-63-3.jpg?t=%E5%A4%8D%E6%95%B0+%E8%AE%BEa%3D%281%2B%E2%88%9A3i%29%2F2%2Cb%3D%281-%E2%88%9A3i%29%2F2%E8%AE%BEa%3D%281%2B%E2%88%9A3i%29%2F2%2Cb%3D%281-%E2%88%9A3i%29%2F2%2C1.%E6%B1%82a%26%23178%3B%2Ca%26%23179%3B%2Cb%26%23178%3B%2Cb%26%23179%3B%E7%9A%84%E5%80%BC+2.%E5%BD%93n%E2%88%88N%2A%E6%97%B6%2C%E8%AE%A1%E7%AE%97a%5En%2Bb%5En%E7%AC%AC%E4%B8%80%E9%97%AE%E6%88%91%E4%BC%9A%E5%81%9Aa%26%23178%3B%3D%28-1%2B%E2%88%9A3i%29%2F2%2Ca%26%23179%3B%3D-1%2Cb%26%23178%3B%3D%28-1-%E2%88%9A3i%29%2F2%2Cb%26%23179%3B%3D-1%E6%B1%82%E8%A7%A3%E7%AC%AC%E4%BA%8C%E9%97%AE%2C%E6%88%91%E7%9A%84%E6%80%9D%E8%B7%AF%3Aa)
复数 设a=(1+√3i)/2,b=(1-√3i)/2设a=(1+√3i)/2,b=(1-√3i)/2,1.求a²,a³,b²,b³的值 2.当n∈N*时,计算a^n+b^n第一问我会做a²=(-1+√3i)/2,a³=-1,b²=(-1-√3i)/2,b³=-1求解第二问,我的思路:a
复数 设a=(1+√3i)/2,b=(1-√3i)/2
设a=(1+√3i)/2,b=(1-√3i)/2,1.求a²,a³,b²,b³的值 2.当n∈N*时,计算a^n+b^n
第一问我会做a²=(-1+√3i)/2,a³=-1,b²=(-1-√3i)/2,b³=-1
求解第二问,我的思路:a+b=1,a²+b²=-1,a³+b³=-2,a^4+b^4=-1,a^5+b^5=1,a^6+b^6=2,a^7+b^7=1,a^8+b^8=-1,a^9+b^9=-2,a^10+b^10=-1.肯定有规律的,我不知道怎么归纳,例如:a^n+b^n=2(-1)^n(n=3k k∈N*)
我观察了一下,当n=1,4,7...n=2,5,8...n=3,6,9......数字没变就是改变符号
复数 设a=(1+√3i)/2,b=(1-√3i)/2设a=(1+√3i)/2,b=(1-√3i)/2,1.求a²,a³,b²,b³的值 2.当n∈N*时,计算a^n+b^n第一问我会做a²=(-1+√3i)/2,a³=-1,b²=(-1-√3i)/2,b³=-1求解第二问,我的思路:a
写成三角函数的形式要好做些吧.
a=cos(π/3)+isin(π/3)
b=cos(-π/3)+isin(-π/3)
所以a^n+b^n=cos(nπ/3)+isin(nπ/3)+cos(-nπ/3)+isin(-nπ/3)=2cos(nπ/3)+isin(nπ/3)-isin(nπ/3)=2cos(nπ/3)
解决方案:多个为a =(1 +√3i的)/ 2 = 1/2 +√3/2i;
=(1 - √3i的)/ 2 = 1/2-√3/2i
为r1 =√[1 ^ 2 +√3)^ 2] = 2。
COSθ(1/2)/ 2 = 1/4,sonθ=(√3/2)/ 2 =√3/4
= R1(COSθ+isinθ)
= 2 (COSI +isinθ)。
...
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解决方案:多个为a =(1 +√3i的)/ 2 = 1/2 +√3/2i;
=(1 - √3i的)/ 2 = 1/2-√3/2i
为r1 =√[1 ^ 2 +√3)^ 2] = 2。
COSθ(1/2)/ 2 = 1/4,sonθ=(√3/2)/ 2 =√3/4
= R1(COSθ+isinθ)
= 2 (COSI +isinθ)。
^ N = [2(COSθ+isinθ)^ N。
= 2 ^ N(cosnθ+isinnθ)
R2 =√[1 ^ 2 +( - √3)^ 2] = 2。
COSθ=(1/2)/ 2 = 1/4,SINθ=( - √3/2)/ 2 = - √3/4。罪(-θ)=√3/4
b = 2时(COSθ+(-isinθ))。
B ^ N = 2 ^ N(cosnθ-isinnθ)
一个^ N + B ^ N = 2 ^(N +1)cosnθ。的n∈N *
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书面形式的三角函数做的更好。 =
余弦(π/ 3)+ ISIN(π/ 3)
B=余弦(-π/ 3)+ ISIN(-π/ 3)
所以一个^ n的+ B ^ n的=余弦(相位偏移nπ/ 3)+ ISIN代码(相位偏移nπ/ 3)+ COS(-相位偏移nπ/ 3)+ ISIN代码(相位偏移nπ/ 3)= 2cos(相位偏移nπ/ 3)+ ISIN代码(相位偏移nπ/ 3)-IS...
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书面形式的三角函数做的更好。 =
余弦(π/ 3)+ ISIN(π/ 3)
B=余弦(-π/ 3)+ ISIN(-π/ 3)
所以一个^ n的+ B ^ n的=余弦(相位偏移nπ/ 3)+ ISIN代码(相位偏移nπ/ 3)+ COS(-相位偏移nπ/ 3)+ ISIN代码(相位偏移nπ/ 3)= 2cos(相位偏移nπ/ 3)+ ISIN代码(相位偏移nπ/ 3)-ISIN代码(相位偏移nπ / 3)= 2cos(相位偏移nπ/ 3)
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