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

  1. Chapter 5 Class 11 Complex Numbers
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Misc, 10 If z1 = 2 โ€“ i, z2 = 1 + i, find |(๐‘ง_1 + ๐‘ง_2 + 1)/(๐‘ง_1 โˆ’ ๐‘ง_2 + 1)| We have to find |(๐‘ง_1 + ๐‘ง_2 + 1)/(๐‘ง_1 โˆ’ ๐‘ง_2 + 1)| First we find (๐‘ง_1 + ๐‘ง_2 + 1)/(๐‘ง_1 โˆ’ ๐‘ง_2 + 1) (๐‘ง_1 + ๐‘ง_2 + 1)/(๐‘ง_1 โˆ’ ๐‘ง_2 + 1) = ("(" 2"โˆ’" ๐‘–")" + "(" 1+ ๐‘–") " + 1)/("(" 2"โˆ’" ๐‘–")" โˆ’" (" 1+ ๐‘–")" + 1) = (2 โˆ’ ๐‘– + 1 + ๐‘– + 1)/(2 โˆ’ ๐‘– โˆ’ 1 โˆ’" " ๐‘– + 1) = (2 + 1 + 1 โˆ’ ๐‘– + ๐‘– )/(2 โˆ’ 1 + 1 โˆ’ ๐‘– โˆ’" " ๐‘–) = (4 + 0)/(2 โˆ’2๐‘– ) = 4/(2 (1 โˆ’ ๐‘–) ) = 2/((1 โˆ’ ๐‘–) ) Rationalizing = 2/(1 โˆ’ ๐‘–) ร— (1 + ๐‘–)/(1 + ๐‘–) = (2 (1 + ๐‘–))/((1 โˆ’ ๐‘–) (1 + ๐‘–)) Using (a โ€“ b) (a + b) = a2 - b2 = (2(1 + ๐‘–))/((1)2 โˆ’ (๐‘–)2) Putting i2 = โˆ’1 = (2 (1 + ๐‘– ))/(1 โˆ’(โˆ’1) ) = (2(1 + ๐‘–))/(1 + 1) = (2 (1 + ๐‘–))/2 = 1 + ๐‘– Hence, (๐‘ง_1 + ๐‘ง_2 + 1)/(๐‘ง_1 โˆ’ ๐‘ง_2 + 1) = 1 + ๐‘– Now we find |(๐‘ง_1 + ๐‘ง_2 + 1)/(๐‘ง_1 โˆ’ ๐‘ง_2 + 1)| i.e. |1 + ๐‘–| Complex number z is of the form ๐‘ฅ + ๐‘– ๐‘ฆ Here x = 1 and y = 1 Modulus of z = |z| = โˆš(๐‘ฅ^2+๐‘ฆ2) = โˆš((1)2+( 1)2) = โˆš(1+1) = โˆš2 So, |(๐‘ง_1 + ๐‘ง_2 + 1)/(๐‘ง_1 โˆ’ ๐‘ง_2 + 1)|= โˆš2 Method 2 To calculate modulus of z We have , z = 1 + ๐‘– Let Polar form of z = r ( cos ฮธ + ๐‘– sin ฮธ ) Form (1) and (2) 1 + ๐‘– (1) = ๐‘Ÿ (cosโกฮธ + ๐‘– sin ฮธ ) 1 + ๐‘– (1) = ๐‘Ÿ cosโกฮธ + ๐‘–r sin ฮธ 1 = r cos ฮธ Squaring both sides (1)2 = (๐‘Ÿ cosฮธ) 1 = r2 cos2 ฮธ r2 cos2 ฮธ = 1 Comparing Imaginary parts 1 = rใ€– sinใ€—โกฮธ Squaring both sides (1)2 = ( r2 sin ฮธ )2 1 = r2 sin2โกฮธ r2 sin2โกฮธ = 1 Adding ( 3 ) and ( 4 ) 1 + 1 = ๐‘Ÿ2 cos2 ฮธ + ๐‘Ÿ2 sin2 ฮธ 2 = ๐‘Ÿ2 (cos2 ฮธ + sin2 ฮธ) 2 = r2 ร— 1 1/2 = r2 1/โˆš2 = r Modulus of ๐‘ง = 1/โˆš2

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