If the complex number `z_1,z_2 and z_3` represent the vertices of an equilateral triangle inscribed in te circel `|z|=2 and z_1=1+isqrt(3)` then

Answer

`z_(2) = -2 , z_(3) = 1 - isqrt3`

Explanation

`z_1=1+ i sqrt(3)=r(cos theta+ i sin theta)" ""[let]"`
`rArr r costheta =1, r sin theta= sqrt(3)`
`rArr " "r=2 and theta= sqrt(3)`
Since `|z_2|=|z_3|=2 " ""[given]"`

Now the triangle `z_1,z_2 and z_3 ` being an equilateral and the sides `z_1z_2 "and " z_1z_3` make an angle `2 pi//3` at the centre.
Therefore `angle POz_2 =(pi)/(3)+(2pi)/(3)= pi`
and `angle POz_3=(pi)/(3)+(2 pi)/(3)+(2pi)/(3)=(5pi)/(3)`
Therefore , `z_2 =2 (cos pi+ i sin pi )=2(-1 + 0 )= -2`
`and z_3=2 (cos""(5pi)/(3)+ isin"(5pi)/3)=2(1/2- i sqrt(3)/2)=1-isqrt(3)`
Alternate Solution
Wherenever vertices of an equilateral triangle having centroid is given its vertices are of the form `z.zomega,zomega^2`
`therefore` If one of the vertex is `z_1=1 +i sqrt(3)` then other two vertices are `(z_1 omega ),(z_1 omega^2)`
`rArr (1+isqrt(3))((-1+isqrt(3)))/2,(1+ isqrt(3))((-1-isqrt(3)))/(2)`
`rArr -((1+3))/2,-((1+i^2(sqrt(3))^2+2isqrt(3)))/2`
`rArr -2,-((-2+2isqrt(3)))/(2)=1 - isqrt(3)`
`therefore z_2 =-2 " and "z_3 =1 - isqrt(3)`