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Misc 10 - If z1 = 2 - i, z2 = 1 + i, find |z1 + z2 + 1| - Miscellaneous

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Misc, 10 If z1 = 2 – i, z2 = 1 + i, find |(𝑧_1 + 𝑧_2 + 1)/(𝑧_1 βˆ’ 𝑧_2 + 1)| We have to find |(𝑧_1 + 𝑧_2 + 1)/(𝑧_1 βˆ’ 𝑧_2 + 1)| First we find (𝑧_1 + 𝑧_2 + 1)/(𝑧_1 βˆ’ 𝑧_2 + 1) (𝑧_1 + 𝑧_2 + 1)/(𝑧_1 βˆ’ 𝑧_2 + 1) = ("(" 2"βˆ’" 𝑖")" + "(" 1+ 𝑖") " + 1)/("(" 2"βˆ’" 𝑖")" βˆ’" (" 1+ 𝑖")" + 1) = (2 βˆ’ 𝑖 + 1 + 𝑖 + 1)/(2 βˆ’ 𝑖 βˆ’ 1 βˆ’" " 𝑖 + 1) = (2 + 1 + 1 βˆ’ 𝑖 + 𝑖 )/(2 βˆ’ 1 + 1 βˆ’ 𝑖 βˆ’" " 𝑖) = (4 + 0)/(2 βˆ’2𝑖 ) = 4/(2 (1 βˆ’ 𝑖) ) = 2/((1 βˆ’ 𝑖) ) Rationalizing = 2/(1 βˆ’ 𝑖) Γ— (1 + 𝑖)/(1 + 𝑖) = (2 (1 + 𝑖))/((1 βˆ’ 𝑖) (1 + 𝑖)) Using (a – b) (a + b) = a2 - b2 = (2(1 + 𝑖))/((1)2 βˆ’ (𝑖)2) Putting i2 = βˆ’1 = (2 (1 + 𝑖 ))/(1 βˆ’(βˆ’1) ) = (2(1 + 𝑖))/(1 + 1) = (2 (1 + 𝑖))/2 = 1 + 𝑖 Hence, (𝑧_1 + 𝑧_2 + 1)/(𝑧_1 βˆ’ 𝑧_2 + 1) = 1 + 𝑖 Now we find |(𝑧_1 + 𝑧_2 + 1)/(𝑧_1 βˆ’ 𝑧_2 + 1)| i.e. |1 + 𝑖| Complex number z is of the form π‘₯ + 𝑖 𝑦 Here x = 1 and y = 1 Modulus of z = |z| = √(π‘₯^2+𝑦2) = √((1)2+( 1)2) = √(1+1) = √2 So, |(𝑧_1 + 𝑧_2 + 1)/(𝑧_1 βˆ’ 𝑧_2 + 1)|= √2 Method 2 To calculate modulus of z We have , z = 1 + 𝑖 Let Polar form of z = r ( cos ΞΈ + 𝑖 sin ΞΈ ) Form (1) and (2) 1 + 𝑖 (1) = π‘Ÿ (cos⁑θ + 𝑖 sin ΞΈ ) 1 + 𝑖 (1) = π‘Ÿ cos⁑θ + 𝑖r sin ΞΈ 1 = r cos ΞΈ Squaring both sides (1)2 = (π‘Ÿ cosΞΈ) 1 = r2 cos2 ΞΈ r2 cos2 ΞΈ = 1 Comparing Imaginary parts 1 = rγ€– sin〗⁑θ Squaring both sides (1)2 = ( r2 sin ΞΈ )2 1 = r2 sin2⁑θ r2 sin2⁑θ = 1 Adding ( 3 ) and ( 4 ) 1 + 1 = π‘Ÿ2 cos2 ΞΈ + π‘Ÿ2 sin2 ΞΈ 2 = π‘Ÿ2 (cos2 ΞΈ + sin2 ΞΈ) 2 = r2 Γ— 1 1/2 = r2 1/√2 = r Modulus of 𝑧 = 1/√2

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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