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

Misc 10 - Chapter 5 Class 11 Complex Numbers - Part 2
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Misc 5 (Method 1) 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) ) = 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 Misc 5 (Method 2) If z1 = 2 – i, z2 = 1 + i, find |(𝑧_1 + 𝑧_2 + 1)/(𝑧_1 βˆ’ 𝑧_2 + 1)| We have , z = 1 + 𝑖 Let Polar form of z = r ( cos ΞΈ + 𝑖 sin ΞΈ ) From (1) and (2) 1 + 𝑖 (1) = π‘Ÿ (cos⁑θ + 𝑖 sin ΞΈ ) 1 + 𝑖 (1) = π‘Ÿ cos⁑θ + 𝑖r sin ΞΈ Comparing real part 1 = r cos ΞΈ Squaring both side (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 2 = r2 √2 = r r = √2 Modulus of 𝑧 = √2

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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.