Modulus, Argument, Polar Representation

Chapter 4 Class 11 Complex Numbers
Serial order wise

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### Transcript

Ex5.2, 5 Convert the given complex number in polar form: – 1 – i Given z = −1− i Let polar form be z = r (cos⁡θ + i sin⁡θ) From (1) & (2) − 1−𝑖 =𝑟 (cos⁡θ+𝑖 sin⁡θ) − 1−𝑖= 𝑟 〖 cos〗⁡θ + 𝑖 r sin⁡θ Adding (3) and (4) 1 + 1 = 𝑟2 cos2 θ+ 𝑟2 sin2θ 2 = 𝑟2 ( cos2 θ+ sin2 θ) 2 = 𝑟2 × 1 2 = 𝑟2 √2 = 𝑟 𝑟 = √2 Finding argument − 1− 𝑖 = r〖 cos〗⁡θ + 𝑖 r sin⁡θ Hence, sin θ = (− 1)/√2& cos θ = (− 1)/√2 Hence, sin θ = (− 1)/√2 & cos θ = (− 1)/√2 Here, sin θ and cos θ both are negative, Hence, θ lies in IIIrd quadrant Argument = – (180° – 45°) = –135° = –135° × 𝜋/180o = ( −3 𝜋)/4 So argument of z = ( −3 𝜋)/4 Hence θ = (−3 𝜋)/4 and r =√2 Polar form of z = r (cos θ + sin θ) = √2 ("cos " ((− 3 𝜋)/4)" – i sin " ((− 3 𝜋)/4))

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.