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复数1-cosx+i*sinx化为指数表达式
人气:155 ℃ 时间:2020-05-22 07:03:48
解答
1-cosx+i*sinx
=2[sin(x/2)]^2+i*2sin(x/2)cos(x/2)
=2sin(x/2)[sin(x/2)+i*cos(x/2)]
=2sin(x/2)[cos(π/2-x/2)+i*sin(π/2-x/2)]
=2sin(x/2)*e^[i*(π/2-x/2)] .
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