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
Solution: Given x = a+b, y = aα +bβ and z = aβ+bα α,β are cube roots of unity. So α = ω, β = ω2 xyz = (a+b)(aα +bβ)( aβ+bα) =... View Article
Solution: Given z2+az+b = 0 Sum of roots, z1+z2 = -a Product of roots , z1z2 = b z2 = z1eiπ/3 z2 = z1( cos π/3 + i sin π/3) = z1(½ +... View Article
Solution: Given complex number = -2√3-2i Let r cos θ = -2√3 r cos θ = -2 Squaring and adding r2(cos2 θ+ sin2 θ) = (-2√3)2+22 r2 = 16 r... View Article
Solution: Since i = eiπ/2 iz = zeiπ/2 Vector iz is obtained by rotating z in anticlockwise direction through 900. |iz| = |i ||z| = 1|z|... View Article
