Question

复数z满足|z-1|=2 argz=7╱6兀求复数z

Answer 1

z=a+bi

argz=7π/6

tan(argz)=b/a=√3/3

a=√3b

|z-1|=2

√[(a-1)^2+b^2]=2

(√3b-1)^2+b^2=4

4b^2-2√3b-3=0

(2b-√3/2)^2=9/4

(2b-√3/2)=-3/2

b=(-3/4+√3/4)

a=-3√3/4+3/4

z=(-3√3/4+3/4) +(-3/4+√3/4) i