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

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Misc 13 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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