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Examples
Example 2 (i)
Example 2 (ii) Important
Example 3
Example 4
Example 5 Important
Example 6 (i)
Example 6 (ii) Important
Example 7 Deleted for CBSE Board 2023 Exams
Example 8 Important Deleted for CBSE Board 2023 Exams
Example 9 Deleted for CBSE Board 2023 Exams
Example 10 Deleted for CBSE Board 2023 Exams
Example 11 Important Deleted for CBSE Board 2023 Exams
Example 12
Example 13 (i) Important
Example 13 (ii)
Example 14 Important
Example 15
Example 16 Important Deleted for CBSE Board 2023 Exams You are here
Last updated at May 29, 2018 by Teachoo
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Example 16 Convert the complex number z = (π β 1)/γcos γβ‘γΟ/3 + π sinβ‘γ Ο/3γ γ in the polar form. Let z = (π β 1)/cosβ‘γ Ο/3 + π sinβ‘γ Ο/3γ γ = (π β 1)/(cosβ‘γ( 180/3 ) + π γsin ( γβ‘γ180/3γ γ ) ) = (π β1)/(cosβ‘γ60Β° + π sinβ‘γ60Β°γ γ ) = (π β1)/(1/2 + β3/2 π) = (π β 1)/( (1 + β3 π)/2) = (2 ( π β1 ))/( 1 + β3 π) Rationalizing = (2 ( π β1))/(1+ β3 π) Γ (1 β β3 π)/(1 β β3 π) = (2 (1 β π) (1 β β3 π))/((1+ β3 π) (1 β β3 π)) = (2 [π (1 β β3 π) β1 (1 β β3 π)])/((1+ β3 π) (1 β β3 π)) = (2[π β β3 π2 β 1 +β3 π])/((1 β β3 π) (1 β β3 π)) Using (a β b) (a + b) = a2 β b2 = (2[ β1 + π + β3 π β β3 π2])/(1^2 β(β3 π)^2 ) = (2 [β1 + π + β3 π β β3 π (π2)])/(1 β 3π2) Putting π2 = - 1 = (2 [β1 + π + β3 π β β3 π (β1 )])/(1 β 3(β1)) = (2 [β1 + π + β3 π+ β3])/(1 + 3) = (2 [ ( β1 + β3 ) + ( π+ β3 π ) ])/4 = (( β1 + β3 )+π ( 1+ β3 ) )/2 = (β3 β1)/2 + π (β3 + 1)/2 Now z = (β3 β1)/2 + π (β3 + 1)/2 Now z = (β3 β1)/2 + π (β3 + 1)/2 Let Polar form be z = r ( cos ΞΈ + π sin ΞΈ ) From (1) and (2) (β3 β1)/2 + π (β3 + 1)/2 = r ( cos ΞΈ + π sin ΞΈ ) (β3 β1)/2 + π (β3 + 1)/2 = r cos ΞΈ + πr sin ΞΈ Adding (3) + (4) (4 β 2β3 )/4 + (4 + 2β3 )/4 = r2 cos2 ΞΈ + r2 sin2 ΞΈ 1/4 ( 4 - 2β3 + 4 + 2β3 ) = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 1/4 ( 4 + 4 - 2β3 + 2β3 ) = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 1/4 ( 8 β 0 ) = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 8/4 = r2 Γ 1 2 = r2 β2 = r r = β2 Now finding argument (β3 β1)/2 + π (β3 + 1)/2 = r cos ΞΈ + πr sin ΞΈ Comparing real part (β3 β1)/2 = r cos ΞΈ Put r = β2 (β3 β1)/2 = β2 cos ΞΈ (β3 β1)/(2β2) = cos ΞΈ Hence, cos ΞΈ = (β3 β1)/(2β2) & sin ΞΈ = (β3 + 1)/(2β2) Hence, cos ΞΈ = (β3 β1)/(2β2) & sin ΞΈ = (β3 + 1)/(2β2) Since sin ΞΈ and cos ΞΈ both Positive Hence ΞΈ lies in the lst Quadrant Argument (ΞΈ ) of z = 75o = 75o Γ π/180 = 5π/12 Hence Polar form of z = r ( cos ΞΈ + i sin ΞΈ ) = β2 ( cos 5π/12 + i sin 5π/12) Hence, Argument = 75Β° = 75 Γ π/180 = 5π/12 Hence , r = β2 , & ΞΈ = 5π/12 Thus, Polar form of z = r ( cos ΞΈ + i sin ΞΈ ) = β2 ( cos 5π/12 + i sin 5π/12)