If z 1 = ( 6 , 3 ) , z 2 = ( 2 , - 1 ) find z 1 z 2 .

Given, z 1 = ( 6 , 3 ) and z 2 = ( 2 , - 1 ) ⇒ z 1 = 6 + 3 i and z 2 = 2 - i Now, z 1 z 2 = 6 + 3 i 2 - i = ( 6 + 3 i ) ( 2 + i ) 4 - i 2 = 12 + 6 i + 6 i + 3 i 2 5 ∵ i 2 = - 1 = 12 + 12 i - 3 5 = 9 + 12 i 5 ⇒ z 1 z 2 = 9 5 + 12 i 5 ∴ z 1 z 2 = 9 5 , 12 5

If z = ( cos θ , sin θ ) , find z - 1 z .

Given, z = cos θ + i sin θ ⇒ 1 z = 1 cos θ + i sin θ ⇒ 1 z = 1 cos θ + i sin θ × cos θ - i sin θ cos θ - i sin θ = cos θ - i sin θ cos 2 θ - i 2 sin 2 θ = cos θ - i sin θ cos 2 θ + sin 2 θ = cos θ - i sin θ Now, z - 1 z = cos θ + i sin θ - cos θ - i sin θ = i sin θ + i sin θ ∴ z - 1 z = 2 i sin θ

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Intermediate Second Year Mathematics Paper 2A>Algebra>Chapter 1 - Complex Numbers>Exercise 1(a)

Andhra Pradesh Board Solutions for Exercise 1: Exercise 1(a)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Exercise 1: Exercise 1(a)

Attempt the free practice questions from Exercise 1: Exercise 1(a) with hints and solutions to strengthen your understanding. Intermediate Second Year Mathematics Paper 2A solutions are prepared by Experienced Embibe Experts.