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Complex Numbers and Quadratic Equations - Identities

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Algebraic identities for complex numbers are identical in form to those for real numbers because complex numbers satisfy the commutative, associative, and distributive laws of algebra.

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For any two complex numbers z1z_1 and z2z_2, the square of their sum follows the identity (z1+z2)2=z12+2z1z2+z22(z_1 + z_2)^2 = z_1^2 + 2z_1z_2 + z_2^2.

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The identity for the difference of two squares, z12−z22=(z1−z2)(z1+z2)z_1^2 - z_2^2 = (z_1 - z_2)(z_1 + z_2), is applicable for complex numbers and is often used in simplification.

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Higher power identities like (z1+z2)3(z_1 + z_2)^3 and (z1−z2)3(z_1 - z_2)^3 also apply to complex numbers, helping in expanding expressions without repeated multiplication.

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When applying these identities, always remember that i2=−1i^2 = -1, i3=−ii^3 = -i, and i4=1i^4 = 1 to simplify the final result into the standard form a+iba + ib.

📐Formulae

(z1+z2)2=z12+2z1z2+z22(z_1 + z_2)^2 = z_1^2 + 2z_1z_2 + z_2^2

(z1−z2)2=z12−2z1z2+z22(z_1 - z_2)^2 = z_1^2 - 2z_1z_2 + z_2^2

(z1+z2)3=z13+3z12z2+3z1z22+z23(z_1 + z_2)^3 = z_1^3 + 3z_1^2z_2 + 3z_1z_2^2 + z_2^3

(z1−z2)3=z13−3z12z2+3z1z22−z23(z_1 - z_2)^3 = z_1^3 - 3z_1^2z_2 + 3z_1z_2^2 - z_2^3

z12−z22=(z1−z2)(z1+z2)z_1^2 - z_2^2 = (z_1 - z_2)(z_1 + z_2)

💡Examples

Problem 1:

Expand and simplify the expression (3+2i)2(3 + 2i)^2 using algebraic identities.

Solution:

Using the identity (z1+z2)2=z12+2z1z2+z22(z_1 + z_2)^2 = z_1^2 + 2z_1z_2 + z_2^2, where z1=3z_1 = 3 and z2=2iz_2 = 2i: (3+2i)2=(3)2+2(3)(2i)+(2i)2(3 + 2i)^2 = (3)^2 + 2(3)(2i) + (2i)^2 =9+12i+4i2= 9 + 12i + 4i^2 Since i2=−1i^2 = -1: =9+12i+4(−1)= 9 + 12i + 4(-1) =9+12i−4= 9 + 12i - 4 =5+12i= 5 + 12i

Explanation:

The square of a binomial identity was applied. The final step involves substituting i2i^2 with −1-1 to combine the real parts.

Problem 2:

Evaluate (1−i)3(1 - i)^3 using the cubic identity.

Solution:

Using the identity (z1−z2)3=z13−3z12z2+3z1z22−z23(z_1 - z_2)^3 = z_1^3 - 3z_1^2z_2 + 3z_1z_2^2 - z_2^3, where z1=1z_1 = 1 and z2=iz_2 = i: (1−i)3=(1)3−3(1)2(i)+3(1)(i)2−(i)3(1 - i)^3 = (1)^3 - 3(1)^2(i) + 3(1)(i)^2 - (i)^3 =1−3i+3i2−i3= 1 - 3i + 3i^2 - i^3 Substitute i2=−1i^2 = -1 and i3=−ii^3 = -i: =1−3i+3(−1)−(−i)= 1 - 3i + 3(-1) - (-i) =1−3i−3+i= 1 - 3i - 3 + i =−2−2i= -2 - 2i

Explanation:

The cubic identity for subtraction was used. Careful substitution of the powers of ii is required to simplify the expression into the a+iba + ib form.

Problem 3:

Factorize the expression z2+16z^2 + 16 using complex number identities.

Solution:

We can rewrite z2+16z^2 + 16 as a difference of squares: z2+16=z2−(−16)z^2 + 16 = z^2 - (-16) Since 16i2=16(−1)=−1616i^2 = 16(-1) = -16: z2+16=z2−(4i)2z^2 + 16 = z^2 - (4i)^2 Now, using the identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b): z2−(4i)2=(z−4i)(z+4i)z^2 - (4i)^2 = (z - 4i)(z + 4i) Therefore, z2+16=(z−4i)(z+4i)z^2 + 16 = (z - 4i)(z + 4i).

Explanation:

The sum of squares a2+b2a^2 + b^2 can be factorized over the set of complex numbers as (a+bi)(a−bi)(a + bi)(a - bi) by expressing it as a difference of squares.

Identities Class 11 Notes & Examples | CBSE Maths