Misc 13 - Find modulus, argument of (1 + 2i)/(1 - 3i) - Miscellaneous

Misc 13 - Chapter 5 Class 11 Complex Numbers - Part 2
Misc 13 - Chapter 5 Class 11 Complex Numbers - Part 3 Misc 13 - Chapter 5 Class 11 Complex Numbers - Part 4 Misc 13 - Chapter 5 Class 11 Complex Numbers - Part 5 Misc 13 - Chapter 5 Class 11 Complex Numbers - Part 6 Misc 13 - Chapter 5 Class 11 Complex Numbers - Part 7 Misc 13 - Chapter 5 Class 11 Complex Numbers - Part 8 Misc 13 - Chapter 5 Class 11 Complex Numbers - Part 9 Misc 13 - Chapter 5 Class 11 Complex Numbers - Part 10

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Question 6 Find the modulus and argument of the complex number ( 1 + 2i)/(1 βˆ’ 3i) . First we solve ( 1 + 2𝑖)/(1 βˆ’ 3𝑖) Let 𝑧 = ( 1 + 2𝑖)/(1 βˆ’ 3𝑖) Rationalizing the same = ( 1 + 2𝑖)/(1 βˆ’ 3𝑖) x ( 1 + 3𝑖)/(1 + 3𝑖) = ((1 + 2𝑖) ( 1 + 3𝑖))/((1 βˆ’ 3𝑖) ( 1 + 3𝑖)) = (1 (1 + 3𝑖) + 2𝑖 (1 + 3𝑖))/((1 βˆ’ 3𝑖) (1 + 3𝑖) ) = ( 1 + 3𝑖+ 2𝑖+ 6𝑖2)/((1 βˆ’ 3𝑖) (1 + 3𝑖) ) Using ( a – b ) ( a + b ) = a2 - b2 = ( 1+ 5𝑖+ 6𝑖2)/((1)2 βˆ’ (3𝑖)2) = (1 + 5𝑖+ 6𝑖2)/(1 βˆ’ 9𝑖2) Putting 𝑖2 = - 1 = (1 + 5𝑖 + 6 (βˆ’1))/(1 βˆ’ 9 (βˆ’1)) = (1 + 5𝑖 βˆ’ 6 )/(1 + 9 ) = ( \βˆ’ 5 + 5𝑖)/10 = ( 5 (βˆ’1 +𝑖))/10 = (βˆ’ 1+ 𝑖)/2 = (βˆ’ 1)/2+𝑖 ( 1/2 ) Thus, 𝑧 = (βˆ’ 1)/2+𝑖 ( 1/2 ) Method 1 To find Modulus Now we have z = (βˆ’ 1)/2 + 𝑖 (1/2) Complex number z is of the form x + 𝑖 y Here x = (βˆ’ 1)/2 and y = 1/2 Modulus of z = |z| = √(π‘₯^2+𝑦2) = √(( (βˆ’ 1)/(2 ))^2+( 1/(2 ))^2 ) = √(1/4+1/4) = √(2/4) = √(1/2) = 1/√2 Modulus of z = 1/√2 Method 2 To calculate modulus of z Given z = (βˆ’1)/2 + 𝑖 (1/2) Let 𝑧=π‘Ÿ (cos⁑θ+𝑖 sinΞΈ) Here r is modulus, and ΞΈ is argument Form (1) and (2) (βˆ’1)/2 + 𝑖 (1/2) = π‘Ÿ (cos⁑θ+𝑖 sinΞΈ) (βˆ’1)/2 + 𝑖 (1/2) = r cos ΞΈ + 𝑖r sin ΞΈ Comparing real part (βˆ’ 1)/2 = r cos ΞΈ Adding (3) and (4) 1/4 + 1/4 = r2 cos2 ΞΈ + r2 sin2 ΞΈ (1+1)/4 = r2 (cos2 ΞΈ+ sin2 ΞΈ) 2/4 = r2 Γ— 1 1/2 = r2 1/√2 = r Modulus of z = 1/√2 Finding argument (βˆ’1)/2 + 𝑖 (1/2) = r cos ΞΈ + 𝑖r sin ΞΈ Hence, sin ΞΈ = 1/√2 & cos ΞΈ = (βˆ’ 1)/√2 Here, sin ΞΈ is positive and cos ΞΈ is negative, Hence, ΞΈ lies in IVth quadrant Argument = 180Β° – 45Β° = 135Β° = 135Β° Γ— πœ‹/180o = ( 3 πœ‹)/4 So argument of z = ( 3 πœ‹)/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.