Lesson 8 of 12 · 7 min
Algebraic identities for complex numbers
NCERT §4.3.7
Asha notices something curious: (2 + i)² comes out as 3 + 4i, the rover's position. Is that a fluke, or do the old algebra identities still work?
The lesson in notes
In short
(z₁ + z₂)² = z₁² + 2z₁z₂ + z₂² for all complex z₁, z₂. The proof expands (z₁ + z₂)(z₁ + z₂) with the distributive law and then uses z₂z₁ = z₁z₂.
In the same way: (z₁ − z₂)² = z₁² − 2z₁z₂ + z₂²; (z₁ + z₂)³ = z₁³ + 3z₁²z₂ + 3z₁z₂² + z₂³; (z₁ − z₂)³ = z₁³ − 3z₁²z₂ + 3z₁z₂² − z₂³; and z₁² − z₂² = (z₁ + z₂)(z₁ − z₂).
Because the laws of addition and multiplication are the same as for real numbers, most identities proved for reals also hold for complex numbers.
Worked example: (2 + i)² = 4 + 4i + i² = 3 + 4i.
Worked example: (1 + 2i)³ = 1 + 3(2i) + 3(2i)² + (2i)³ = 1 + 6i − 12 − 8i = −11 − 2i. Check by steps: (1 + 2i)² = −3 + 4i, and (−3 + 4i)(1 + 2i) = −3 − 6i + 4i − 8 = −11 − 2i.
Useful results: (1 + i)² = 2i and (1 − i)² = −2i, so (1 + i)⁴ = (2i)² = −4 and (1 + i)⁸ = 16.