If the arguments of 1 - i 3 + i 1 + 3 i and Z - 2 Z ¯ - 1 are equal, then the locus of Z is part of a circle with centre a , b . The value of 1 a + b is

Step-by-step solution

Let, Z = x + i y Now, Z - 2 Z ¯ - 1 = Z Z ¯ - 2 Z ¯ - Z + 2 = x 2 + y 2 - 3 x + 2 + i y argument of 1 - i 3 + i 1 + 3 i = - π 4 + π 6 + π 3 = π 4 Now, since both arguments are equal ⇒ x 2 + y 2 - 3 x + 2 = y   ( where ,   y > 0 ) ⇒ x 2 + y 2 - 3 x - y + 2 = 0 Hence, the centre of the circle is 3 2 , 1 2 = a , b ⇒ 1 a + b = 1 3 2 + 1 2 = 1 2 =0. 5