Last updated at Aug. 27, 2021 by
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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
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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 Deleted for CBSE Board 2022 Exams
Example 8 Important Deleted for CBSE Board 2022 Exams
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Example 10
Example 11 Important
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Example 13 (i) Important Deleted for CBSE Board 2022 Exams
Example 13 (ii) Deleted for CBSE Board 2022 Exams You are here
Example 14 Important
Example 15
Example 16 Important Deleted for CBSE Board 2022 Exams
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