复数计算:(1/2+(根号3/2i))^5+(-1/2+(根号3/2i))^3
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复数计算:(1/2+(根号3/2i))^5+(-1/2+(根号3/2i))^3
![复数计算:(1/2+(根号3/2i))^5+(-1/2+(根号3/2i))^3](/uploads/image/z/5689667-11-7.jpg?t=%E5%A4%8D%E6%95%B0%E8%AE%A1%E7%AE%97%3A%281%2F2%2B%28%E6%A0%B9%E5%8F%B73%2F2i%29%29%5E5%2B%28-1%2F2%2B%28%E6%A0%B9%E5%8F%B73%2F2i%29%29%5E3)
答:把原式化简成三角形式. 利用 [r(cosθ+isinθ)]^n=(r^n)[cosnθ+isinnθ] 公式计算
[cos(π/3)+sin(π/3)i]^5+[cos(2/3π)+sin(2/3π)i]³
=[cos(5Xπ/3)+sin(5Xπ/3]+[cos(3X2/3π)+sin(3X2/3π)i]
=[cos(2π-π/3)+cos(2π)]+[sin(2π-π/3)+sin(2π)]i
=【cos(π/3)-1】+[-sin(π/3)-0]i
=(1/2-1)-(√3/2)i
=-(1/2+√3/2)i
[cos(π/3)+sin(π/3)i]^5+[cos(2/3π)+sin(2/3π)i]³
=[cos(5Xπ/3)+sin(5Xπ/3]+[cos(3X2/3π)+sin(3X2/3π)i]
=[cos(2π-π/3)+cos(2π)]+[sin(2π-π/3)+sin(2π)]i
=【cos(π/3)-1】+[-sin(π/3)-0]i
=(1/2-1)-(√3/2)i
=-(1/2+√3/2)i
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