【求助】高二选修数学题

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.

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
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.