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Ex 5.2, 1 Find the modulus and the argument of the complex number z = −1 − i√3 Given z = − 1 − 𝑖√3 Let z = r (𝒄𝒐𝒔⁡𝜽 + 𝒊 𝒔𝒊𝒏⁡𝜽) Here, r is modulus, and θ is argument Comparing (1) & (2) − 1 − 𝑖 √3 = r (cos⁡θ + 𝑖 sin⁡θ) − 1 − 𝒊 √𝟑 = r〖 𝒄𝒐𝒔〗⁡𝜽 + 𝒊 r 𝒔𝒊𝒏⁡𝜽 Comparing real and imaginary parts Comparing Real Parts −𝟏 = 𝒓 𝒄𝒐𝒔⁡𝜽 Squaring both sides (−1)2 = ( r〖 cos〗⁡θ)2 1 = r2〖 cos"2" 〗⁡θ r2〖 𝐜𝐨𝐬"2" 〗⁡𝜽 = 1 Comparing Imaginary Parts –√𝟑 = r〖 𝒔𝒊𝒏〗⁡𝜽 Squaring both sides ("– " √3)^2 = r2 sin2 θ – √3 × – √3 = r2 sin2⁡θ 3 = r2 sin2⁡θ r2 𝒔𝒊𝒏𝟐⁡𝜽 = 3 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 𝒓𝟐 = 𝟒 𝑟 =± √4 𝑟 = ±2 Since modulus is always positive Hence, Modulus = r = 2 Finding argument Given −1 − 𝑖 √3 = r〖 cos〗⁡θ + 𝑖 r sin⁡θ Putting r = 2 −1 − 𝑖 √3 = 2〖 cos〗⁡θ + 𝑖 2 sin⁡θ Comparing Real Parts −𝟏 =𝟐 𝒄𝒐𝒔⁡𝜽 −1/2 =𝑐𝑜𝑠⁡𝜃 𝒄𝒐𝒔⁡𝜽=(−𝟏)/𝟐 Comparing Imaginary Parts –√𝟑 = 2〖 𝒔𝒊𝒏〗⁡𝜽 (−√3)/2 = 〖 𝑠𝑖𝑛〗⁡𝜃 〖 𝒔𝒊𝒏〗⁡𝜽 = (−√𝟑)/𝟐 Hence, sin 𝛉 = (−√𝟑)/𝟐 & cos 𝛉 = (−𝟏)/𝟐 Since both sin θ and cos θ are negative , Argument will be in 3rd quadrant Since −𝜋 ≤ Argument ≤ 𝜋 Argument = – (𝜋 −𝜋/3) = (−𝟐𝝅)/𝟑

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.