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设复数z满足4z+2*z的共轭复数=3倍根号3+i,w=sina-icosa,求z的值和|z-w|的取值范围.
要有具体过程·注明:a代表的是角度
人气:166 ℃ 时间:2020-04-11 03:49:14
解答
z=c+bi,
4(c+bi)+2(c-bi) = 6c+2bi=3*3^(1/2)+i,
c=3^(1/2)/2,b=1/2.
z = 3^(1/2)/2 + i/2.
z-w=3^(1/2)/2 + i/2 -sina+icosa = [3^(1/2)/2 -sina] + i[1/2 + cosa]
|z-w|^2 = [3^(1/2)/2 - sina]^2 + [1/2 + cosa]^2
= 3/4 -3^(1/2)sina + 1/4 + cosa + 1
= 2 + 2[cosa*1/2-sina*3^(1/2)/2]
= 2 + 2cos[a+PI/3]
0 <= |z-w|^2 = 2 + 2cos[a+PI/3] <= 4,
0 <= |z-w| <= 2.
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