If z 1 = - 1 and z 2 = i , then find Arg z 1 z 2

Given, z 1 = - 1 and z 2 = i We know that, argument of purely real number with negative real part is π . Argument of purely imaginary part with positive imaginary part is π 2 . Now, Arg z 1 z 2 = Arg z 1 - Arg z 2 = π - π 2 = π 2 ∴ Arg z 1 z 2 = 90 °

If x = cis θ , then find the value of x 6 + 1 x 6 .

Given that, x = cis θ = cos θ + i sin θ = e i θ Now consider, x 6 + 1 x 6 = e 6 i θ + 1 e 6 i θ = e 6 i θ + e - 6 i θ ∵ 1 x = x - 1 = cos 6 θ + i sin 6 θ + cos - 6 θ + i sin - 6 θ = cos 6 θ + i sin 6 θ + cos 6 θ - i sin 6 θ ∵ cos - θ = cos θ a n d sin - θ = - sin θ = 2 cos 6 θ Therefore, x 6 + 1 x 6 = 2 cos 6 θ .

Show that the points in the Argand diagram represented by the complex numbers 2 + 2 i , - 2 - 2 i , - 2 3 + 2 3 i are the vertices of an equilateral triangle.

Given, 2 + 2 i , - 2 - 2 i , - 2 3 + 2 3 i Let A ( 2 , 2 ) , B ( - 2 , - 2 ) , C ( - 2 3 , 2 3 ) be points represents given complex numbers in the Argand plane. A B = ( 2 + 2 ) 2 + ( 2 + 2 ) 2 ⇒ A B = 4 2 B C = ( - 2 + 2 3 ) 2 + ( - 2 - 2 3 ) 2 ⇒ B C = 4 + 12 - 8 3 + 4 + 12 + 8 3 ⇒ B C = 4 2 A C = ( 2 + 2 3 ) 2 + ( 2 - 2 3 ) 2 ⇒ A C = 4 + 12 + 8 3 + 4 + 12 - 8 3 ⇒ A C = 4 2 ∴ A B = A C = B C So, △ A B C is equilateral.

If Z = ( cos ⁡ θ , sin ⁡ θ ) , then find Z − 1 Z

We have, Z = ( cos ⁡ θ , sin ⁡ θ ) i.e. Z = cos ⁡ θ + i sin ⁡ θ Let f Z = Z - 1 Z ⇒ f Z = cos ⁡ θ + i sin ⁡ θ - 1 cos ⁡ θ + i sin ⁡ θ ⇒ f Z = e i θ - 1 e i θ [Using Euler's formula] ⇒ f Z = e i θ - e - i θ ⇒ f Z = 2 i sin θ ∵ sin ⁡ A = e j A − e − i A 2 i

If x + i y = 1 1 + cos θ + i sin θ , then show that 4 x 2 - 1 = 0 .

Given, x + i y = 1 1 + cos θ + i sin θ ⇒ x + i y = 1 + cos θ - i sin θ 1 + cos θ + i sin θ 1 + cos θ - i sin θ ⇒ x + i y = 1 + cos θ - i sin θ ( 1 + cos θ ) 2 - i 2 sin 2 θ ⇒ x + i y = 1 + cos θ - i sin θ 1 + 2 cos θ + cos 2 θ - i 2 sin 2 θ ⇒ x + i y = 1 + cos θ - i sin θ 1 + 2 cos θ + cos 2 θ + sin 2 θ ⇒ x + i y = 1 + cos θ - i sin θ 1 + 2 cos θ + 1 ⇒ x + i y = 1 + cos θ 2 + 2 cos θ - i sin θ 2 + 2 cos θ Then, x = 1 + cos θ 2 + 2 cos θ ⇒ x = 1 + cos θ 2 ( 1 + cos θ ) ⇒ x = 1 2 ⇒ 2 x = 1 ⇒ 4 x 2 = 1 ∴ 4 x 2 - 1 = 0 Hence, proved.

If n is an integer and Z = cis ⁡ θ , θ ≠ ( 2 n + 1 ) π 2 then show that Z 2 n − 1 Z 2 n + 1 = i tan ⁡ n θ .

Given that : Z = cis θ = cos θ + i sin θ , θ ≠ ( 2 n + 1 ) π 2 , where n is an integer. To prove: Z 2 n - 1 Z 2 n + 1 = i tan n θ From, LHS, we have, = Z 2 n - 1 Z 2 n + 1 = ( cos θ + i sin θ ) 2 n - 1 ( cos θ + i sin θ ) 2 n + 1 = cos 2 n θ + i sin 2 n θ - 1 cos 2 n θ + i sin 2 n θ + 1 ∵ ( cos A + i sin A ) n = cos n A + i sin n A , n ∈ Z = - ( 1 - cos 2 n θ ) + i sin 2 n θ ( 1 + cos 2 n θ ) + i sin 2 n θ = i 2 2 sin 2 n θ + 2 i sin n θ · cos n θ 2 cos 2 n θ + 2 i sin n θ · cos n θ ∵ - 1 = i 2 , 1 + cos 2 A = 2 cos 2 A , 1 - cos 2 A = 2 sin 2 A , sin 2 A = 2 sin A cos A = 2 i sin n θ { cos n θ + i sin n θ } 2 cos n θ { cos n θ + i sin n θ } = i tan n θ = RHS Hence, proved.

If z = 2 - i 7 then, show that 3 z 3 - 4 z 2 + z + 88 = 0

Given, z = 2 - i 7 . . . i To Prove: 3 z 3 - 4 z 2 + z + 88 = 0 Equation i can be written as z - 2 = - i 7 ⇒ ( z - 2 ) 2 = ( - i 7 ) 2 ⇒ z 2 - 4 z + 4 = - 7 ∵ i 2 = - 1 ⇒ z 2 - 4 z + 11 = 0 . . . . i i Multiplying with 3 z on both sides, ⇒ 3 z 3 - 12 z 2 + 33 z = 0 ⇒ 3 z 3 - 4 z 2 + z + 88 + - 8 z 2 + 32 z - 88 = 0 ⇒ 3 z 3 - 4 z 2 + z + 88 - 8 z 2 - 4 z + 11 = 0 ∴ 3 z 3 - 4 z 2 + z + 88 = 0 [From i i ] Hence, proved.

If x = cis θ , then find the value of x 6 + 1 x 6 .

Given that, x = cis θ = cos θ + i sin θ = e i θ Now consider, x 6 + 1 x 6 = e 6 i θ + 1 e 6 i θ = e 6 i θ + e - 6 i θ ∵ 1 x = x - 1 = cos 6 θ + i sin 6 θ + cos - 6 θ + i sin - 6 θ = cos 6 θ + i sin 6 θ + cos 6 θ - i sin 6 θ ∵ cos - θ = cos θ a n d sin - θ = - sin θ = 2 cos 6 θ Therefore, x 6 + 1 x 6 = 2 cos 6 θ .

If z 1 = - 1 and z 2 = i , then find Arg z 1 z 2

Given, z 1 = - 1 and z 2 = i We know that, argument of purely real number with negative real part is π . Argument of purely imaginary part with positive imaginary part is π 2 . Now, Arg z 1 z 2 = Arg z 1 - Arg z 2 = π - π 2 = π 2 ∴ Arg z 1 z 2 = 90 °

Find the value of ( 1 - i ) 8 .

Given, ( 1 - i ) 8 = ( 1 - i ) 2 4 = 1 - 2 i + i 2 4 = 1 - 2 i - 1 4 = - 2 i 4 = 16 i 4 = 16 - 1 2 ∵ ( 1 - i ) 8 = 16

Express the complex number - 1 - i 3 in modulus amplitude form.

Let, - 1 - i 3 = r cos θ + r i sin θ Comparing real and imaginary parts, we get r cos θ = - 1 ...(i) and r sin θ = - 3 ...(ii) Here, r is the modulus of the given complex number ⇒ r = | - 1 - i 3 | = 1 + ( - 3 ) 2 = 1 + 3 = 4 = 2 Now, ( i i ) ( i ) = tan θ = 3 ⇒ θ = 4 π 3 , since - 1 - i 3 = - 1 , - 3 is in III quadrant. ∴ - 1 - i 3 = 2 cos 4 π 3 + i sin 4 π 3

Show that the points in the Argand diagram represented by the complex numbers 2 + 2 i , - 2 - 2 i , - 2 3 + 2 3 i are the vertices of an equilateral triangle.

Given, 2 + 2 i , - 2 - 2 i , - 2 3 + 2 3 i Let A ( 2 , 2 ) , B ( - 2 , - 2 ) , C ( - 2 3 , 2 3 ) be points represents given complex numbers in the Argand plane. A B = ( 2 + 2 ) 2 + ( 2 + 2 ) 2 ⇒ A B = 4 2 B C = ( - 2 + 2 3 ) 2 + ( - 2 - 2 3 ) 2 ⇒ B C = 4 + 12 - 8 3 + 4 + 12 + 8 3 ⇒ B C = 4 2 A C = ( 2 + 2 3 ) 2 + ( 2 - 2 3 ) 2 ⇒ A C = 4 + 12 + 8 3 + 4 + 12 - 8 3 ⇒ A C = 4 2 ∴ A B = A C = B C So, △ A B C is equilateral.

If n is an integer and Z = cis ⁡ θ , θ ≠ ( 2 n + 1 ) π 2 then show that Z 2 n − 1 Z 2 n + 1 = i tan ⁡ n θ .

Given that : Z = cis θ = cos θ + i sin θ , θ ≠ ( 2 n + 1 ) π 2 , where n is an integer. To prove: Z 2 n - 1 Z 2 n + 1 = i tan n θ From, LHS, we have, = Z 2 n - 1 Z 2 n + 1 = ( cos θ + i sin θ ) 2 n - 1 ( cos θ + i sin θ ) 2 n + 1 = cos 2 n θ + i sin 2 n θ - 1 cos 2 n θ + i sin 2 n θ + 1 ∵ ( cos A + i sin A ) n = cos n A + i sin n A , n ∈ Z = - ( 1 - cos 2 n θ ) + i sin 2 n θ ( 1 + cos 2 n θ ) + i sin 2 n θ = i 2 2 sin 2 n θ + 2 i sin n θ · cos n θ 2 cos 2 n θ + 2 i sin n θ · cos n θ ∵ - 1 = i 2 , 1 + cos 2 A = 2 cos 2 A , 1 - cos 2 A = 2 sin 2 A , sin 2 A = 2 sin A cos A = 2 i sin n θ { cos n θ + i sin n θ } 2 cos n θ { cos n θ + i sin n θ } = i tan n θ = RHS Hence, proved.