Modulus, Argument, Polar Representation

Chapter 4 Class 11 Complex Numbers
Serial order wise

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Transcript

Ex5.2, 2 Find the modulus and the argument of the complex number π§ = β β3 + π Method (1) To calculate modulus of z z = - β3 + π Complex number z is of the form x + πy Where x = - β3 and y = 1 Modulus of z = |z| = β(π₯^2+π¦^2 ) = β(( β β3 )2+( 1 )2 ) = β(3+1) = β4 = 2 Hence |z| = 2 Modulus of z = 2 Method (2) to calculate Modulus of z Given z = β β3 + π Let z = r (cosβ‘ΞΈ + π sinβ‘ΞΈ) Here r is modulus, and ΞΈ is argument From (1) & (2) β β3 + π = r (cosβ‘ΞΈ+π sinβ‘ΞΈ ) β β3 + π = rγ cosγβ‘ΞΈ + π r sinβ‘ΞΈ Comparing Real parts β3 = r cosβ‘ΞΈ Squaring both sides (β3)^2 = (π cosβ‘ΞΈ)2 3 = π2 γ cos"2" γβ‘ΞΈ Adding (3) & (4) 3 + 1 = r2 cos2β‘ΞΈ + r2 sin2β‘ΞΈ 4 = π2 cos2β‘ΞΈ + r2 sin2β‘ΞΈ 4 = π2 ( cos2β‘ΞΈ + sin2β‘ΞΈ ) 4 = π2 Γ 1 4 = π2 β4 = π r = 2 Hence, Modulus = 2 Finding argument β β3 + π = rγ cosγβ‘ΞΈ + π r sinβ‘ΞΈ Comparing real part ββ3 = r cosβ‘ΞΈ Putting r = 2 ββ3 = 2cosβ‘ΞΈ β β3/2 = cosβ‘ΞΈ cosβ‘ΞΈ = β β3/2 Hence, sinβ‘ΞΈ = 1/2 & cos ΞΈ = ββ3/2 Since sin ΞΈ is positive and cos ΞΈ is negative , Argument will be in IInd quadrant Argument = 180Β° β 30Β° = 150Β° = 150 Γ π/180 = (5 π)/3