Show that the points in the Argand plane represented by the complex numbers - 2 + 7 i , - 3 2 + 1 2 i , 4 - 3 i , 7 2 1 + i are the vertices of a rhombus.

Given, - 2 + 7 i , - 3 2 + 1 2 i , 4 - 3 i , 7 2 1 + i Let the four points represented by the complex numbers be A - 2 , 7 , B - 3 2 , 1 2 , C 4 , - 3 , D 7 2 , 7 2 The distance between the given points are, A B = - 2 + 3 2 2 + 7 - 1 2 2 = 1 4 + 169 4 = 170 2 B C = 4 + 3 2 2 + - 3 - 1 2 2 = 121 4 + 49 4 = 170 2 C D = 4 - 7 2 2 + - 3 - 7 2 2 = 1 4 + 169 4 = 170 2 A D = - 2 - 7 2 2 + 7 - 7 2 2 = 121 4 + 49 4 = 170 2 Slope of A C = 7 + 3 - 2 - 4 = 10 - 6 = - 5 3 Slope of B D = 7 2 - 1 2 7 2 + 3 2 = 3 5 Hence, A C and B D are perpendicular. ∴ A B C D is a rhombus.

The complex number z has argument θ , 0 < θ < π 2 and satisfy the equation | z - 3 i | = 3 . Then prove that cot θ - 6 z = i .

Let, z = x + i y Given, | z - 3 i | = 3 ⇒ | x + i y - 3 i | = 3 ⇒ | x + i ( y - 3 ) | = 3 ⇒ x 2 + ( y - 3 ) 2 = 3 ⇒ x 2 + y 2 - 6 y + 9 = 9 ⇒ x 2 + y 2 - 6 y = 0 ⇒ x 2 + y 2 = 6 y ⇒ 6 y x 2 + y 2 = 1 ⇒ 6 y | z | 2 = 1 ⇒ 6 y z z = 1 ⇒ 6 z = z ¯ y ⇒ 6 z = x - i y y ⇒ 6 z = x y - i ⇒ 6 z = cot θ - i ∴ cot θ - 6 z = i Hence, proved.