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

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

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Transcript

Example 8 Convert the complex number ( 16 )/(1 + 3) into polar form. Let z = ( 16)/(1 + 3) Rationalizing = ( 16)/(1 + 3) (1 3)/(1 3) = ( 16 ( 1 3 ))/(" " (1 + 3) (1 3)) Using ( a b ) ( a + b ) = a2 - b2 = ( 16 ( 1 3 ))/(( 1 )2 ( 3)2) = ( 16 ( 1 3 ))/( 1 3 2) Putting 2 = 1 = ( 16 ( 1 3 ))/(1 3 ( 1)) = ( 16 ( 1 3 ))/(1 + 3 ) = ( 16 ( 1 3 ))/4 = 4 ( 1 3 ) = 4 4 ( 3 ) = 4 + 4 3 Hence, z = 4 + 4 3 Hence, z = 4 + 4 3 Let polar form be z = r ( cos + sin ) From (1) and (2) 4 + 4 3 = r ( cos + sin ) 4 + 4 3 = r cos + sin Comparing real part 4 = r cos Squaring both sides ( 4)2 = ( r cos )2 16 = r2 cos2 Adding (3) and (4) 16 + 48 = r2 cos2 + r2 sin2 64 = r2 ( cos2 + sin2 ) 64 = r2 1 64 = r2 64 = r 8 = r r = 8 Finding argument 4 + 4 3 = r cos + sin Comparing real part 4 = r cos Put r = 8 - 4 = 8 cos ( 4)/8 = cos ( 1)/2 = cos cos = ( 1)/2 Hence, sin = 3/2 & cos =( 1)/2 Hence, sin = 3/2 & cos =( 1)/2 Since sin is positive and cos is negative Argument will be in IInd quadrant Argument = 180 60 = 120 = 120 /180 = 2 /3 Hence = 2 /3 and r = 8 Polar form of z = r ( cos + sin ) = 8 ( cos 2 /3 + i sin 2 /3)

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.