If n is a positive integer, prove that 1 + sin θ − i cos θ 1 + sin θ + i cos θ n = cos n π 2 − θ − i sin n π 2 − θ .

To prove 1 + sin θ − i cos θ 1 + sin θ + i cos θ n = cos n π 2 − θ − i sin n π 2 − θ Let Z = sin θ - i cos θ . So Z = 1 Now, 1 Z = 1 sin θ - i cos θ = sin θ + i cos θ sin θ - i cos θ sin θ + i cos θ = sin θ + i cos θ sin 2 θ + cos 2 θ = sin θ + i cos θ LHS = 1 + sin θ - i cos θ 1 + sin θ + i cos θ n = 1 + Z 1 + 1 Z n = Z Z + 1 1 + Z n = Z n And Z n = sin θ - i cos θ n = cos π 2 − θ − i sin π 2 − θ n = cos n π 2 − θ − i sin n π 2 − θ = RHS [By De Moivre's Theorem, cos θ + i sin θ n = cos n θ + i sin n θ for all n ∈ Z ] Hence, proved.

' P ' represents the variable complex number z . Find the locus of P . If Re z - 1 z + i = 1 .

Let, z = x + i y Therefore, z - 1 z + i = x + i y - 1 x + i y + i = ( x - 1 ) + i y x + i ( y + 1 ) × x - i ( y + 1 ) x - i ( y + 1 ) [Multiply and divide with conjugate of the denominator] = x ( x - 1 ) + i x y - i x - 1 y + 1 - i 2 y y + 1 x 2 - i 2 ( y + 1 ) 2 = x ( x - 1 ) + y ( y + 1 ) + i ( x y ) - ( x - 1 ) ( y + 1 ) x 2 + ( y + 1 ) 2 [Using i 2 = - 1 ] Now, Re z - 1 z + 1 = 1 ⇒ x ( x - 1 ) + y ( y + 1 ) x 2 + ( y + 1 ) 2 = 1 ⇒ x 2 + y 2 - x + y = x 2 + y 2 + 2 y + 1 ⇒ - x + y = 2 y + 1 ∴ Locus of P is x + y + 1 = 0 .

If n is a positive integer, prove that 1 + sin θ − i cos θ 1 + sin θ + i cos θ n = cos n π 2 − θ − i sin n π 2 − θ .

To prove 1 + sin θ − i cos θ 1 + sin θ + i cos θ n = cos n π 2 − θ − i sin n π 2 − θ Let Z = sin θ - i cos θ . So Z = 1 Now, 1 Z = 1 sin θ - i cos θ = sin θ + i cos θ sin θ - i cos θ sin θ + i cos θ = sin θ + i cos θ sin 2 θ + cos 2 θ = sin θ + i cos θ LHS = 1 + sin θ - i cos θ 1 + sin θ + i cos θ n = 1 + Z 1 + 1 Z n = Z Z + 1 1 + Z n = Z n And Z n = sin θ - i cos θ n = cos π 2 − θ − i sin π 2 − θ n = cos n π 2 − θ − i sin n π 2 − θ = RHS [By De Moivre's Theorem, cos θ + i sin θ n = cos n θ + i sin n θ for all n ∈ Z ] Hence, proved.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Complex Numbers>Tamilnadu Board-2019>Q 7

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12th Tamil Nadu Board

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If $- x - i y$ lies in the first quadrant, then $- i x + y$ lies in the:

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Important Questions on Complex Numbers

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12th Tamil Nadu Board

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Complex Numbers>Tamilnadu Board-2019>Q 8

If $z_{1} = 1 + 2 i, z_{2} = 1 - 3 i$ and $z_{3} = 2 + 4 i$ then, the points on the Argand diagram representing $z_{1} z_{2} z_{3}, 2 z_{1} z_{2} z_{3}, - 7 z_{1} z_{2} z_{3}$ are: