Solution: Let z = x+iy iz = -y+ix z+iz = (x+iy+i(x+iy) = (x-y)+i(x+y) So the vertices of the triangle are (x,y), (-y, x) and (x-y, x+y)... View Article
Solution: Given modulus = 2 Argument is 2π/3. r = |z| = 2 x = r cos θ y = r sin θ x = 2 cos 2π/3 = 2(-½) = -1 y = 2 sin 2π/3 = 2(√3/2) =... View Article
Solution: Let z = x+iy Given |z-i |+ |z+i| = k |x+iy-i |+ |x+iy+i| = k Taking modulus √(x2+(y-1)2)+ √(x2+(y+1)2) = k √(x2+(y-1)2) =... View Article