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

Misc 5 - Chapter 5 Class 11 Complex Numbers - Part 2
Misc 5 - Chapter 5 Class 11 Complex Numbers - Part 3 Misc 5 - Chapter 5 Class 11 Complex Numbers - Part 4 Misc 5 - Chapter 5 Class 11 Complex Numbers - Part 5 Misc 5 - Chapter 5 Class 11 Complex Numbers - Part 6 Misc 5 - Chapter 5 Class 11 Complex Numbers - Part 7 Misc 5 - Chapter 5 Class 11 Complex Numbers - Part 8

 

 

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Misc 5 Convert the following in the polar form: (i) (1 + 7𝑖)/(2 − 𝑖)2 Let z = ( 1+7𝑖)/(2 − 𝑖)2 Using ( a – b)2 = a2 + b2 – 2ab = ((1 + 7𝑖))/((2)2+ (i)2 − 2 × 2 × 𝑖) = ( 1 + 7𝑖)/(4 + 𝑖2− 4𝑖) Putting 𝑖2 = -1 = (1 + 7𝑖)/(4 + ( −1 ) − 4𝑖) = (1 + 7𝑖 )/(4 − 1 − 4𝑖 ) = (1 + 7𝑖)/(3 − 4𝑖) Rationalizing the Same = (1 + 7𝑖)/(3 − 4𝑖) × (3 + 4𝑖)/(3 + 4𝑖) = ((1+7𝑖) (3 + 4𝑖 ))/((3 −4𝑖) (3 −4𝑖)) = (1 (3+4𝑖) +7𝑖 (3 + 4𝑖 ))/((3 −4𝑖) (3 + 4𝑖)) = (3 + 4𝑖 + 21𝑖 + 28𝑖2)/((3 − 4𝑖) (3 + 4𝑖)) = (3 + 25𝑖 + 28𝑖2)/((3 − 4𝑖) (3 + 4𝑖)) Using (a – b) (a + b) = a2 – b2 = (3 + 25𝑖 + 28𝑖2)/((3)2− (4𝑖)2) = (3 + 25𝑖 + 28𝑖2)/(9 − 16𝑖2) Putting 𝑖2 = 1 = (3 + 25𝑖 + 28 (−1 ))/(9 −16 ( −1 )) = (3 + 25𝑖 − 28)/(9 + 16) = (3 − 28 + 25𝑖)/25 = (− 25 + 25𝑖)/25 = ( 25 ( −1 + 𝑖 ))/25 = - 1 + 𝑖 Hence, z = – 1 + 𝑖 Let polar form be z = 𝑟 (cos⁡θ+𝑖sin⁡θ ) From (1) and (2) –1 + 𝑖 = r (cos θ + 𝑖 sin θ) –1 + 𝑖 = r cos θ + 𝑖 r sin θ Comparing real part − 1 = r cos θ Squaring both sides (− 1 )2 =( 𝑟 cos⁡θ )^2 1 = 𝑟2 cos2θ Adding (3) and (4) 1 + 1 = 𝑟2 cos2 θ + 𝑟2 sin2 θ 1 + 1 = r2 cos2 θ + r2 sin2 θ 2 = r2 ( cos2 θ + sin2 θ ) 2 = r2 × 1 2 = r2 √2 = r r = √2 Now finding argument –1 + 𝑖 = r cos θ + 𝑖 r sin θ Comparing real part − 1 = r cos θ Putting r =√2 − 1 = √2 cos θ (− 1 )/√2 = cos θ cos θ = (− 1 )/√2 Hence, cos θ = (− 1 )/√2 & sin θ = ( 1)/√2 Hence, cos θ = (− 1 )/√2 & sin θ = ( 1)/√2 Since, sin θ is positive and cos θ is negative, Hence, θ lies in IInd quadrant Argument = 180° – 45° = 135° = 135° × 𝜋/(180°) = 3𝜋/4 Hence, argument of 𝑧 = 3𝜋/4 Hence 𝑟 = √2 and θ = 3𝜋/4 Polar form of 𝑧=𝑟 (cos⁡θ+sin⁡θ ) = √2 (cos(3𝜋/4)+sin(3𝜋/4))

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.