Example 8 - Convert -16/1 + i root3 into polar form - Polar representation

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
Ask Download

Transcript

Example 8 Convert the complex number (βˆ’16 )/(1 + π‘–βˆš3) into polar form. Let z = (βˆ’ 16)/(1 + π‘–βˆš3) Rationalizing = (βˆ’ 16)/(1 + 𝑖 √3) Γ— (1 βˆ’ 𝑖 √3)/(1 βˆ’ 𝑖 √3) = (βˆ’ 16 ( 1 βˆ’ 𝑖 √3 ))/(" " (1 + 𝑖 √3) (1 βˆ’ 𝑖 √3)) Using ( a – b ) ( a + b ) = a2 - b2 = (βˆ’ 16 ( 1 βˆ’ 𝑖 √3 ))/(( 1 )2 βˆ’(𝑖 √3)2) = (βˆ’ 16 ( 1 βˆ’ 𝑖 √3 ))/( 1βˆ’3𝑖2) Putting 𝑖2 = – 1 = (βˆ’ 16 ( 1 βˆ’ 𝑖 √3 ))/(1βˆ’3 (βˆ’1)) = (βˆ’ 16 ( 1 βˆ’ 𝑖 √3 ))/(1 + 3 ) = (βˆ’ 16 ( 1 βˆ’ 𝑖 √3 ))/4 = – 4 ( 1 βˆ’ 𝑖 √3 ) = – 4 – 4 (βˆ’ 𝑖 √3 ) = – 4 + 𝑖4 √3 Hence, z = – 4 + 𝑖4 √3 Hence, z = – 4 + 𝑖4 √3 Let polar form be z = r ( cosΞΈ + 𝑖 sinΞΈ ) From (1) and (2) – 4 + 𝑖4 √3 = r ( cosΞΈ + 𝑖 sinΞΈ ) – 4 + 𝑖4 √3 = r cosΞΈ + 𝑖 π‘ŸsinΞΈ Comparing real part – 4 = r cos ΞΈ Squaring both sides (– 4)2 = ( r cos ΞΈ )2 16 = r2 cos2 ΞΈ Adding (3) and (4) 16 + 48 = r2 cos2 ΞΈ + r2 sin2 ΞΈ 64 = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 64 = r2 Γ— 1 64 = r2 √64 = r 8 = r r = 8 Finding argument – 4 + 𝑖4 √3 = r cosΞΈ + 𝑖 π‘ŸsinΞΈ Comparing real part – 4 = r cos ΞΈ Put r = 8 - 4 = 8 cos ΞΈ (βˆ’ 4)/8 = cos ΞΈ (βˆ’ 1)/2 = cos ΞΈ cos ΞΈ = (βˆ’ 1)/2 Hence, sin ΞΈ= √3/2 & cos ΞΈ=(βˆ’ 1)/2 Hence, sin ΞΈ= √3/2 & cos ΞΈ=(βˆ’ 1)/2 Since sin ΞΈ is positive and cos ΞΈ is negative Argument will be in IInd quadrant Argument = 180Β° – 60Β° = 120Β° = 120 Γ— πœ‹/180 = 2πœ‹/3 Hence ΞΈ = 2πœ‹/3 and r = 8 Polar form of z = r ( cos ΞΈ + 𝑖 sin ΞΈ ) = 8 ( cos 2πœ‹/3 + i sin 2πœ‹/3)

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.