Complex Number Operations

[MathDefinition title="Complex Numbers"] A complex number is of the form a + ib, where a and b are real numbers and i = √(−1). Here, a is the real part and b is the imaginary part. [/MathDefinition]

[MathDefinition title="Modulus of a Complex Number"] The modulus of a complex number z = a + ib is |z| = √(a² + b²). It represents the distance of the point (a, b) from the origin. [/MathDefinition]

[MathKeyFormula title="Operations on Complex Numbers"] If z₁ = a + ib and z₂ = c + id:

1. Addition: z₁ + z₂ = (a + c) + i(b + d)
2. Subtraction: z₁ − z₂ = (a − c) + i(b − d)
3. Multiplication: z₁ × z₂ = (ac − bd) + i(ad + bc)
4. Division: z₁/z₂ = (z₁ × z̄₂)/(z₂ × z̄₂) where z̄₂ is conjugate of z₂
5. Conjugate: z̄ = a − ib
6. |z|² = z × z̄ = a² + b² [/MathKeyFormula]

[MathNote title="Properties of Complex Numbers"]

1. The sum of a complex number and its conjugate is real: z + z̄ = 2a
2. The difference is purely imaginary: z − z̄ = 2ib
3. The product of a complex number and its conjugate is real: z × z̄ = |z|²
4. |z₁z₂| = |z₁||z₂| and |z₁/z₂| = |z₁|/|z₂| [/MathNote]

[MathExample title="Example 1: Addition and Subtraction"] Problem: If z₁ = 3 + 2i and z₂ = 1 − 4i, find z₁ + z₂ and z₁ − z₂.

Solution: z₁ + z₂ = (3 + 2i) + (1 − 4i) = (3 + 1) + i(2 − 4) = 4 − 2i

z₁ − z₂ = (3 + 2i) − (1 − 4i) = (3 − 1) + i(2 − (−4)) = 2 + 6i

Therefore, z₁ + z₂ = 4 − 2i and z₁ − z₂ = 2 + 6i [/MathExample]

[MathExample title="Example 2: Multiplication"] Problem: Find the product of (2 + 3i) and (4 − i).

Solution: (2 + 3i)(4 − i) = 2(4) + 2(−i) + 3i(4) + 3i(−i) = 8 − 2i + 12i − 3i² = 8 + 10i − 3(−1) [since i² = −1] = 8 + 10i + 3 = 11 + 10i

Therefore, (2 + 3i)(4 − i) = 11 + 10i [/MathExample]

[MathExample title="Example 3: Division"] Problem: Find (3 + 2i)/(1 − i).

Solution: Multiply numerator and denominator by the conjugate of the denominator:

(3 + 2i)/(1 − i) = (3 + 2i)(1 + i)/((1 − i)(1 + i))

Numerator: (3 + 2i)(1 + i) = 3 + 3i + 2i + 2i² = 3 + 5i − 2 = 1 + 5i

Denominator: (1 − i)(1 + i) = 1 − i² = 1 − (−1) = 2

Therefore, (3 + 2i)/(1 − i) = (1 + 5i)/2 = 1/2 + 5i/2 [/MathExample]

[MathExample title="Example 4: Finding Modulus"] Problem: Find the modulus of z = 3 − 4i and verify that |z|² = z × z̄.

Solution: z = 3 − 4i |z| = √(3² + (−4)²) = √(9 + 16) = √25 = 5

z̄ = 3 + 4i

z × z̄ = (3 − 4i)(3 + 4i) = 9 + 12i − 12i − 16i² = 9 + 16 = 25

|z|² = 5² = 25

Therefore, |z| = 5 and |z|² = z × z̄ = 25 [/MathExample]