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📅 Year 2024
Q1CBSE Board Exam 20243 Marks
Find the modulus and argument of the complex number z = −1 − i√3, and write it in polar form.
▶ Show Detailed Solution
|z| = √[(−1)² + (−√3)²] = √(1 + 3) = √4 = 2.
Since both are negative, θ lies in QIII: tan α = √3/1 = √3 ⇒ α = 60°.
θ = 180° + 60° = 240° = 4π/3.
|z| = 2, arg z = 4π/3, z = 2(cos 4π/3 + i sin 4π/3)
Q2CBSE Board Exam 20242 Marks
Express (1 + i)⁴ in the form a + ib.
▶ Show Detailed Solution
(1 + i)² = 1 + 2i + i² = 1 + 2i − 1 = 2i.
(1 + i)⁴ = (2i)² = 4i² = −4.
(1 + i)⁴ = −4 + 0i
📅 Year 2023
Q1CBSE Board Exam 20233 Marks
If x + iy = (1 + i)/(1 − i), prove that x² + y² = 1.
▶ Show Detailed Solution
(1 + i)/(1 − i) × (1 + i)/(1 + i) = (1 + i)²/(1² − i²) = (1 + 2i + i²)/2 = (2i)/2 = i.
So x + iy = i ⇒ x = 0, y = 1.
x² + y² = 0 + 1 = 1 (proved)