Examples

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Example 7 Deleted for CBSE Board 2023 Exams You are here

Example 8 Important Deleted for CBSE Board 2023 Exams

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Example 10 Deleted for CBSE Board 2023 Exams

Example 11 Important Deleted for CBSE Board 2023 Exams

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Example 16 Important Deleted for CBSE Board 2023 Exams

Last updated at May 29, 2018 by Teachoo

Example 7 Convert the given complex number in polar form: 1 + i√3. Given z = 1+ √3i Let polar form be z = r (cosθ + i sinθ) From ( 1 ) & ( 2 ) 1 + √3i = r ( cosθ + i sinθ) 1 + √3i = r〖 cos〗θ + 𝑖 r sinθ Adding (3) & (4) 1 + 3 = r2 cos2θ + r2 sin2θ 4 = 𝑟2 cos2θ + r2 sin2θ 4 = 𝑟2 ( cos2θ + sin2θ ) 4 = 𝑟2 × 1 4 = 𝑟2 √4 = 𝑟 r = 2 Hence, Modulus = 2 Finding argument 1 + √3i = r〖 cos〗θ + 𝑖 r sinθ Comparing real part 1 = r cosθ Putting r = 2 1 = 2cosθ 1/2 = cosθ cosθ = 1/2 Hence, sinθ = √3/2 & cos θ = 1/2 Hence, sinθ = √3/2 & cos θ = 1/2 Since sin θ and cos θ both are positive, Argument will be in Ist quadrant Argument = 60° = 60 × 𝜋/180 = 𝜋/3 Hence θ = 𝜋/3 and r = 2 Polar form of z = r ( cos θ + 𝑖 sin θ ) = 2 ( cos 𝜋/3 + 𝑖 sin 𝜋/3)