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Example 2 (i)
Example 2 (ii) Important
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Example 5 Important
Example 6 (i)
Example 6 (ii) Important
Example 7
Example 8 Important
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Example 11 Important
Example 12
Example 13 (i) Important
Example 13 (ii) You are here
Example 14 Important
Example 15
Example 16 Important
Last updated at Aug. 27, 2021 by Teachoo
Example, 13 Find the modulus and argument of the complex numbers: (ii) 1/(1 + π) First we simplify 1/(1 + π) 1/(1 + π) Rationalizing = 1/(1 + π) Γ (1 β π)/(1 β π) = (1 Γ (1 β π))/(" " (1 + π)(1 β π) ) Using (a β b) (a + b) = a2 β b2 = (1 β π)/((1)^2 β(π)^2 ) = (1 βπ)/(1 β(β1) ) = (1 β π)/(1 + 1 ) = (1 β π)/2 = 1/2 + π ((β 1)/2) Now z = 1/2 + π ((β 1)/2) We calculate modulus by two different methods Method 1 To calculate Modulus of z z = 1/2 + π ( (β 1)/2 ) Complex number z is of the form π₯ + ππ¦ Here π₯ = 1/2 and π¦ = (β 1)/2 Modulus of z = |z| = β(π₯^2+π¦2) = β(( 1/2 )^2+( (β 1)/2 )^2 ) = β( 1/4+1/4 ) = β( (1 + 1)/4 ) = β( 2/4 ) = β( 1/(" " 2" " )) = 1/(" " β( 2) " " ) β Modulus of π§ is 1/(" " β( 2) " " ) Method 2 to calculate Modulus of z Given π§ = 1/2 + π ( (β 1)/2 ) Let z = π(πππ β‘ΞΈ+π sin ΞΈ) Here r is modulus, and ΞΈ is argument Form (1) and (2) 1/2 + π ( (β 1)/2 ) = π(πππ β‘ΞΈ+π sin ΞΈ) 1/2 + π ( (β 1)/2 ) = rcos ΞΈ + π r sin ΞΈ Adding (3) and (4) 1/4 + 1/4 = r2 cos2 ΞΈ + r2 sin2 ΞΈ (1 + 1)/4 = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 2/4 = π2 (cos2 ΞΈ+sin2 ΞΈ) 1/2 = π2 Γ 1 β(1/2) = r 1/β2 = π π = 1/β2 β Modulus = 1/β2 Finding argument 1/2 + π ( (β 1)/2 ) = rcos ΞΈ + π r sin ΞΈ Comparing real part 1/2 = r cos ΞΈ Put r = 1/β2 1/2 = 1/β2 cos ΞΈ β2/2 = cos ΞΈ 1/β2 = cos ΞΈ β cos ΞΈ = 1/β2 Hence, cos ΞΈ = 1/β2 & sin ΞΈ = (β 1)/β2 Here, sin ΞΈ is negative and cos ΞΈ is positive, Hence, ΞΈ lies in IVth quadrant So, Argument = β 45Β° = β 45Β° Γ π/(180Β°) = (β π)/4 Hence, argument of π§ = (β π)/4