Misc 5 - Convert in polar form: (i) (1 + 7i)/(2 - i)2 - Polar representation

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Misc 5 Convert the following in the polar form: (i) ( 1 + 7𝑖)/(2 βˆ’ 𝑖)2 Let z = ( 1+7𝑖)/(2 βˆ’ 𝑖)2 Using ( a – b)2 = a2 + b2 – 2ab = ((1 + 7𝑖))/((2)2+ (i)2 βˆ’ 2 Γ— 2 Γ— 𝑖) = ( 1 + 7𝑖)/(4 + 𝑖2βˆ’ 4𝑖) Putting 𝑖2 = -1 = (1 + 7𝑖)/(4 + ( βˆ’1 ) βˆ’ 4𝑖) = (1 + 7𝑖 )/(4 βˆ’ 1 βˆ’ 4𝑖 ) = (1 + 7𝑖)/(3 βˆ’ 4𝑖) Rationalizing the Same = (1 + 7𝑖)/(3 βˆ’ 4𝑖) Γ— (3 + 4𝑖)/(3 + 4𝑖) = ((1+7𝑖) (3 + 4𝑖 ))/((3 βˆ’4𝑖) (3 βˆ’4𝑖)) = (1 (3+4𝑖) +7𝑖 (3 + 4𝑖 ))/((3 βˆ’4𝑖) (3 + 4𝑖)) = (3 + 4𝑖 + 21𝑖 + 28𝑖2)/((3 βˆ’ 4𝑖) (3 + 4𝑖)) = (3 + 25𝑖 + 28𝑖2)/((3 βˆ’ 4𝑖) (3 + 4𝑖)) Using (a – b) (a + b) = a2 – b2 = (3 + 25𝑖 + 28𝑖2)/((3)2βˆ’ (4𝑖)2) = (3 + 25𝑖 + 28𝑖2)/(9 βˆ’ 16𝑖2) Putting 𝑖2 = 1 = (3 + 25𝑖 + 28 (βˆ’1 ))/(9 βˆ’16 ( βˆ’1 )) = (3 + 25𝑖 βˆ’ 28)/(9 + 16) = (3 βˆ’ 28 + 25𝑖)/25 = (βˆ’ 25 + 25𝑖)/25 = ( 25 ( βˆ’1 + 𝑖 ))/25 = - 1 + 𝑖 Hence, z = – 1 + 𝑖 Let polar form be z = π‘Ÿ (cos⁑θ+𝑖sin⁑θ ) From (1) and (2) –1 + 𝑖 = r (cos ΞΈ + 𝑖 sin ΞΈ) –1 + 𝑖 = r cos ΞΈ + 𝑖 r sin ΞΈ Comparing real part βˆ’ 1 = r cos ΞΈ Squaring both sides (βˆ’ 1 )2 =( π‘Ÿ cos⁑θ )^2 1 = π‘Ÿ2 cos2ΞΈ Adding (3) and (4) 1 + 1 = π‘Ÿ2 cos2 ΞΈ + π‘Ÿ2 sin2 ΞΈ 1 + 1 = r2 cos2 ΞΈ + r2 sin2 ΞΈ 2 = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 2 = r2 Γ— 1 2 = r2 √2 = r r = √2 Now finding argument –1 + 𝑖 = r cos ΞΈ + 𝑖 r sin ΞΈ Comparing real part βˆ’ 1 = r cos ΞΈ Putting r =√2 βˆ’ 1 = √2 cos ΞΈ (βˆ’ 1 )/√2 = cos ΞΈ cos ΞΈ = (βˆ’ 1 )/√2 Hence, cos ΞΈ = (βˆ’ 1 )/√2 & sin ΞΈ = ( 1)/√2 Hence, cos ΞΈ = (βˆ’ 1 )/√2 & sin ΞΈ = ( 1)/√2 Since, sin ΞΈ is positive and cos ΞΈ is negative, Hence, ΞΈ lies in IInd quadrant Argument = 180Β° – 45Β° = 135Β° = 135Β° Γ— πœ‹/(180Β°) = 3πœ‹/4 Hence, argument of 𝑧 = 3πœ‹/4 Hence π‘Ÿ = √2 and ΞΈ = 3πœ‹/4 Polar form of 𝑧=π‘Ÿ (cos⁑θ+sin⁑θ ) = √2 (cos(3πœ‹/4)+sin(3πœ‹/4)) Misc 5 Convert the following in the polar form: (ii) ( 1 + 3𝑖)/(1 βˆ’ 2𝑖) Let z = ( 1 + 3𝑖)/(1 βˆ’ 2𝑖) Rationalizing = (1 + 3𝑖)/(1 βˆ’ 2𝑖) Γ— (1 + 2𝑖)/(1 + 2𝑖) = ((1 + 3𝑖) (1 + 2𝑖 ))/((1 βˆ’ 2𝑖) (1 + 2𝑖)) = (1 (1 + 2𝑖) + 3𝑖 (1 + 2𝑖 ))/((1 βˆ’ 2𝑖) (1 + 2𝑖)) = (1 +2𝑖 +3𝑖 +6𝑖2)/((1 βˆ’ 2𝑖) (1 + 2𝑖)) = (1 + 5𝑖 +6𝑖2)/((1 βˆ’2𝑖) (1 +2𝑖)) Using (a – b) (a + b) = a2 – b2 = (1 + 5𝑖 +6𝑖2)/((1)2βˆ’ (2𝑖)2) = (1 + 5𝑖 +6𝑖2)/(1 βˆ’4𝑖2) Putting 𝑖2 = –1 = (1 + 5𝑖 +6 (βˆ’1 ))/(1 βˆ’4 ( βˆ’1 )) = (1 + 5𝑖 βˆ’6)/(1 +4) = (1 βˆ’6 + 5𝑖)/5 = (βˆ’ 5 + 5𝑖)/5 = (βˆ’ 5 ( βˆ’1 + 𝑖 ))/5 = - 1 + 𝑖 Hence, z = – 1 + 𝑖 Let polar form be z = π‘Ÿ (cos⁑θ+𝑖sin⁑θ ) From (1) and (2) –1 + 𝑖 = r (cos ΞΈ + 𝑖 sin ΞΈ) –1 + 𝑖 = r cos ΞΈ + 𝑖 r sin ΞΈ Comparing real part βˆ’ 1 = r cos ΞΈ Squaring both sides (βˆ’ 1 )2 =( π‘Ÿ cos⁑θ )^2 1 = π‘Ÿ2 cos2ΞΈ Adding (3) and (4) 1 + 1 = π‘Ÿ2 cos2 ΞΈ + π‘Ÿ2 sin2 ΞΈ 1 + 1 = r2 cos2 ΞΈ + r2 sin2 ΞΈ 2 = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 2 = r2 Γ— 1 2 = r2 √2 = r r = √2 Now finding argument –1 + 𝑖 = r cos ΞΈ + 𝑖 r sin ΞΈ Comparing real part βˆ’ 1 = r cos ΞΈ Putting r =√2 βˆ’ 1 = √2 cos ΞΈ (βˆ’ 1 )/√2 = cos ΞΈ cos ΞΈ = (βˆ’ 1 )/√2 Hence, cos ΞΈ = (βˆ’ 1 )/√2 & sin ΞΈ = ( 1)/√2 Hence, cos ΞΈ = (βˆ’ 1 )/√2 & sin ΞΈ = ( 1)/√2 Since, sin ΞΈ is positive and cos ΞΈ is negative, Hence, ΞΈ lies in IInd quadrant Argument = 180Β° – 45Β° = 135Β° = 135Β° Γ— πœ‹/(180Β°) = 3πœ‹/4 Hence, argument of 𝑧 = 3πœ‹/4 Hence π‘Ÿ = √2 and ΞΈ = 3πœ‹/4 Polar form of 𝑧=π‘Ÿ (cos⁑θ+sin⁑θ ) = √2 (cos(3πœ‹/4)+sin(3πœ‹/4))

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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