How do you find z if arg(z+2) = pi/3 and arg (z-2) = 5pi/6?

Answer 1

let z=x+iy then z+2=(x+2)+iy and z-2=(x-2)+iy
tan (arg(z+2))=y/(x+2) and tan arg (z-2)=y/(x-2)
so y/x+2=tan pi/3 and y/x-2=tan 5pi/6
divide and get x-2/x+2=-3
so x-2=-3x-6 so x=-1
so y=sqrt3
so the number is (-1+isqrt3) (i is sqrt(-1))