若[3/(1/2-√3/2i)]^n∈R,则正整数n的最小值是

若[3/(1/2-√3/2i)]^n∈R,则正整数n的最小值是

分子分母同乘以(1/2+√3/2i)，得：
原式=[3*(1/2+√3/2i)]^n
={3*[sin(π/6)+icos(π/6)]}^n
根据复数的乘方运算法则
原式=(3^n)*[sin(nπ/6)+icos(nπ/6)]
因为上式∈R，所以虚部3^n*cos(nπ/6)=0
nπ/6=kπ+π/2，k是任意整数
n=6k+3
因为n是正整数，所以n的最小值为3

正整数n的最小值是3

3/[(1/2)-(√3/2)i]=3[(1/2)+(√3/2)i]

{3[(1/2)+(√3/2)i]}²=9[(-1/2)+(√3/2)i]

{3[(1/2)+(√3/2)i]}³
=27[(1/2)+(√3/2)i][(-1/2)+(√3/2)i]
=27[(-1/4)+(√3/4)i-(√3/4)i-3/4]
=-27

z=3/[1/2-(√3/2)i]
=6/(1-√3i )
=6(1+√3i )/4
=(3/2)(1+√3i )
arcz = arctan (√3/1) = π/3
|z|^2 =(9/4)(1+3)=9
|z| =3

z = 3(cos(π/3) + isin(π/3) )
z^n ∈R
min n = 3追问z = 3(cos(π/3) + isin(π/3) )
z^n ∈R
min n = 3
最后这几步怎么判断n=3，看不太懂

∵3/(1/2-√3/2i)=6/(1-√3i)=(3/2)(1+√3i)
现要使[3/(1/2-√3/2i)]^n∈R也即是要(1+√3i)^n∈R
当n=1时，原式=(3/2)(1+√3i)∉R
当n=2时，原式=(9/4)(√3i+1)²=(9/2)(√3i-1)
当n=3时，原式=(27/8)(1+√3i)³=-27
∴满足题意的最小正整数n=3