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Question 7 - Polar representation - Chapter 4 Class 11 Complex Numbers
Last updated at April 16, 2024 by Teachoo
Chapter 4 Class 11 Complex Numbers
Concept wise
- Quadaratic equation
- Two complex numbers equal
- Convert to a + ib form
- Identities (square, cube of 2 complex numbers)
- Division
- Power of i(odd and even)
- Conjugate
- Modulus,argument
-
Polar representation
- Proof- Replacing iota with -iota
- Proof- Taking general complex numbers
- Proof- Solving
- Proof- Using modulus conjugate property (Pg 102)
Transcript
Question 7 Convert the given complex number in polar form: 3 + i. Given z = 3 + i Let polar form be z = r (cos + i sin ) From (1) & (2) 3 + i = r (cos + i sin ) 3 + = r cos + r sin Adding (3) & (4) 3 + 1 = r2 cos2 + r2 sin2 4 = 2 cos2 + r2 sin2 4 = 2 ( cos2 + sin2 ) 4 = 2 1 4 = 2 4 = r = 2 Hence, Modulus = 2 Finding argument 3 + = r cos + r sin Comparing real part 3 = r cos Putting r = 2 3 = 2cos 3/2 = cos cos = 3/2 Hence, sin = 1/2 & cos = 3/2 Hence, sin = 1/2 & cos = 3/2 Since sin and cos both are positive, Argument will be in Ist quadrant Argument = 30 = 30 /180 = /6 Hence = /6 and r = 2 Polar form of z = r ( cos + sin ) = 2 ( cos /6 + sin /6)
