Given z 2 + ( 4 3 ) z + 13 = 0 On an Argand diagram with origin O , show the position vectors O A → and O B → representing the roots of the equation. Describe the geometrical relationship between O A → and O B → .

Given z 2 + ( 4 3 ) z + 13 = 0 ⇒ z 2 + 2 × 2 3 z + 12 + 13 - 12 = 0 ⇒ z + 2 3 2 = - 1 ⇒ z + 2 3 = - 1 ⇒ z + 2 3 = ± i ⇒ z = - 2 3 ± i Given that position vectors O A → and O B → representing the roots of the equation. Hence we can represent the numbers z 1 = - 2 3 + i and z 2 = - 2 3 - i by the position vectors O A → = - 2 3 1 and O B → = - 2 3 - 1 . Hence the argand plane is shown below. From the complex plane, it can be seen that both vectors start from origin to the respective point and one vector is exactly a reflection of the other. Therefore both vectors are reflections in the real axis.

Given z 2 + ( 4 3 ) z + 13 = 0 Find the modulus and argument of each root.

Given z 2 + ( 4 3 ) z + 13 = 0 ⇒ z 2 + 2 × 2 3 z + 12 + 13 - 12 = 0 ⇒ z + 2 3 2 = - 1 ⇒ z + 2 3 = - 1 ⇒ z + 2 3 = ± i ⇒ z = - 2 3 ± i It has roots z 1 = - 2 3 - i and z 2 = - 2 3 + i Modulus of a complex number x + i y is given by x 2 + y 2 . The Argument of a complex number a + i b is given by tan - 1 b a . Use the principal argument of a complex number by observing the angle made with the positive x - axis. For root z 1 = - 2 3 - i Modulus of z 1 = ( - 2 3 ) 2 + ( - 1 ) 2 = 12 + 1 = 13 arg ( - 2 3 - i ) = tan - 1 - 1 - 2 3 = 2 . 86 (in radians) For root z 2 = - 2 3 + i Modulus of z 2 = ( - 2 3 ) 2 + 1 2 = 12 + 1 = 13 arg ( - 2 3 + i ) = tan - 1 - 1 2 3 = - tan - 1 1 2 3 = - 2 . 86 (in radians) ∴ z 1 = z 2 = 13 , arg z 1 = 2 . 86 and arg z 2 = - 2 . 86

z = 4 3 - 4 i Given that w = 2 2 cos π 12 + i sin π 12 , write z w in the form r e i θ , where r > 0 and - π < θ ⩽ π .

Given z = 4 3 - 4 i Modulus of a complex number x + i y is given by x 2 + y 2 . The Argument of a complex number a + i b is given by tan - 1 b a . Use the principal argument of a complex number by observing the angle made with the positive x - axis. For root z = 4 3 - 4 i Modulus of z = ( 4 3 ) 2 + ( 4 ) 2 ⇒ z = 16 × 3 + 16 ⇒ z = 64 = 8 ⇒ a r g ( 4 3 - 4 i ) = tan - 1 - 4 4 3 ⇒ tan - 1 - 1 3 = - π 6 ∴ z = 8 , a r g z = - π 6 . Euler's relationship: r e i θ = r ( cos θ + i sin θ ) where r is the modulus of z and θ is the a r g ( z ) . So z = 4 3 - 4 i can be expressed as z = 8 ( cos - π 6 + i sin - π 6 ) Given w = 2 2 cos π 12 + i sin π 12 ⇒ z w = 8 ( cos - π 6 + i sin - π 6 ) 2 2 cos π 12 + i sin π 12 ⇒ z w = 2 2 ( cos - π 6 + i sin - π 6 ) cos π 12 + i sin π 12 ⇒ z w = 2 2 e i - π 6 e i π 12 ⇒ z w = 2 2 e i - π 6 × e i - π 12 ⇒ z w = 2 2 e i - π 6 - π 12 ⇒ z w = 2 2 e - i π 4