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Example 7 - Convert in polar form: 1 + i root 3 - NCERT - Polar representation

a Example 7 - Chapter 5 Class 11 Complex Numbers - Part 2
Example 7 - Chapter 5 Class 11 Complex Numbers - Part 3 Example 7 - Chapter 5 Class 11 Complex Numbers - Part 4

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Example 7 Convert the given complex number in polar form: 1 + i√3. Given z = 1+ √3i Let polar form be z = r (cos⁡θ + i sin⁡θ) From ( 1 ) & ( 2 ) 1 + √3i = r ( cos⁡θ + i sin⁡θ) 1 + √3i = r〖 cos〗⁡θ + 𝑖 r sin⁡θ Adding (3) & (4) 1 + 3 = r2 cos2⁡θ + r2 sin2⁡θ 4 = 𝑟2 cos2⁡θ + r2 sin2⁡θ 4 = 𝑟2 ( cos2⁡θ + sin2⁡θ ) 4 = 𝑟2 × 1 4 = 𝑟2 √4 = 𝑟 r = 2 Hence, Modulus = 2 Finding argument 1 + √3i = r〖 cos〗⁡θ + 𝑖 r sin⁡θ Comparing real part 1 = r cos⁡θ Putting r = 2 1 = 2cos⁡θ 1/2 = cos⁡θ cos⁡θ = 1/2 Hence, sin⁡θ = √3/2 & cos θ = 1/2 Hence, sin⁡θ = √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)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.