source: ncert
topic: algebra
class_level: class_11
chapter: ch5
difficulty: intermediate
Class 11 | Ch5: Complex Numbers and Quadratic Equations COMPLEX NUMBER: z=a+ib where a=real part, b=imaginary part, i=sqrt(-1). i^1=i, i^2=-1, i^3=-i, i^4=1. Pattern repeats every 4. i^n: divide n by 4, use remainder (0→1, 1→i, 2→-1, 3→-i). ALGEBRA OF COMPLEX NUMBERS: Addition: (a+ib)+(c+id)=(a+c)+i(b+d). Multiplication: (a+ib)(c+id)=(ac-bd)+i(ad+bc). Conjugate of z=a+ib is z_bar=a-ib. |z|=sqrt(a^2+b^2) (modulus/absolute value). z×z_bar=|z|^2=a^2+b^2. Division: z1/z2=z1×z2_bar/|z2|^2. ARGAND PLANE: Complex number a+ib plotted as point (a,b). POLAR FORM: z=r(cos(theta)+i sin(theta)) where r=|z|, theta=arg(z). arg(z)=tan^(-1)(b/a) (adjust for quadrant). DE MOIVRE'S THEOREM: (cos(theta)+i sin(theta))^n=cos(ntheta)+i sin(ntheta). CUBE ROOTS OF UNITY: x^3=1. Solutions: 1, omega, omega^2 where omega=(-1+isqrt3)/2. Properties: 1+omega+omega^2=0. omega^3=1. omega_bar=omega^2. QUADRATIC WITH COMPLEX ROOTS: ax^2+bx+c=0, D=b^2-4ac<0: roots=(-b±isqrt(-D))/2a. Roots are conjugates. SOLVED EXAMPLES: Example 1: Modulus and argument of z=-1+i*sqrt3. |z|=sqrt(1+3)=2. tan(alpha)=sqrt3/1=sqrt3 → alpha=pi/3. z is in Q2 (real negative, imag positive). arg(z)=pi-pi/3=2pi/3. Polar form: z=2(cos(2pi/3)+i sin(2pi/3)). Example 2: Find (1+i)^10 using De Moivre. 1+i=sqrt2(cos(pi/4)+i sin(pi/4)). (1+i)^10=(sqrt2)^10(cos(10pi/4)+i sin(10pi/4))=32(cos(5pi/2)+i sin(5pi/2))=32(0+i)=32i. Example 3: Cube roots of unity. 1+omega+omega^2=0. omega^3=1. If 1,omega,omega^2 are roots: sum=0=-b/a → b=0. Product=1=c/a → c=a. Equation: x^3-1=0. COMMON MISTAKES: i^2=-1, NOT +1. i^0=1 (not i). Conjugate changes sign of imaginary part only: conjugate of (3-2i) is (3+2i). |z|^2=a^2+b^2 (always positive real number).