Check sibling questions

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

Β 

This video is only available for Teachoo black users

Β 

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

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))

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.