Miscellaneous

Chapter 4 Class 11 Complex Numbers
Serial order wise

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Misc 5 (Method 1) If z1 = 2 β i, z2 = 1 + i, find |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)| We have to find |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)| First we find (π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1) (π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1) = ("(" 2"β" π")" + "(" 1+ π") " + 1)/("(" 2"β" π")" β" (" 1+ π")" + 1) = (2 β π + 1 + π + 1)/(2 β π β 1 β" " π + 1) = (2 + 1 + 1 β π + π )/(2 β 1 + 1 β π β" " π) = (4 + 0)/(2 β2π ) = 4/(2 (1 β π) ) = 4/(2 (1 β π) ) = 2/((1 β π) ) Rationalizing = 2/(1 β π) Γ (1 + π)/(1 + π) = (2 (1 + π))/((1 β π) (1 + π)) Using (a β b) (a + b) = a2 - b2 = (2(1 + π))/((1)2 β (π)2) Putting i2 = β1 = (2 (1 + π ))/(1 β(β1) ) = (2(1 + π))/(1 + 1) = (2 (1 + π))/2 = 1 + π Hence, (π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1) = 1 + π Now we find |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)| i.e. |1 + π| Complex number z is of the form π₯ + π π¦ Here x = 1 and y = 1 Modulus of z = |z| = β(π₯^2+π¦2) = β((1)2+( 1)2) = β(1+1) = β2 So, |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)|= β2 Misc 5 (Method 2) If z1 = 2 β i, z2 = 1 + i, find |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)| We have , z = 1 + π Let Polar form of z = r ( cos ΞΈ + π sin ΞΈ ) From (1) and (2) 1 + π (1) = π (cosβ‘ΞΈ + π sin ΞΈ ) 1 + π (1) = π cosβ‘ΞΈ + πr sin ΞΈ Comparing real part 1 = r cos ΞΈ Squaring both side (1)2 = (π cosΞΈ) 1 = r2 cos2 ΞΈ r2 cos2 ΞΈ = 1 Comparing Imaginary parts 1 = rγ sinγβ‘ΞΈ Squaring both sides (1)2 = ( r2 sin ΞΈ )2 1 = r2 sin2β‘ΞΈ r2 sin2β‘ΞΈ = 1 Adding (3) and (4) 1 + 1 = π2 cos2 ΞΈ + π2 sin2 ΞΈ 2 = π2 (cos2 ΞΈ + sin2 ΞΈ) 2 = r2 Γ 1 2 = r2 β2 = r r = β2 Modulus of π§ = β2