Find all complex numbers z such that \(|z-1|=|z+3|=|z-i|\).
Express each answer in the form a+bi , where a and b are real numbers.
+23252 Let z = a + bi
Then:
1) |z - 1| = | (a + bi) - 1 | = | (a - 1) + bi | = sqrt( (a - 1)2 + (b)2 )
2) |z + 3| = | (a + bi) + 3 | = | (a + 3) + bi | = sqrt( (a + 3)2 + b2 )
3) |z - i| = | (a + bi) - i | = | a + (b - 1)i | = sqrt( (a)2 + (b - 1)2 }
Setting 1) and 2) equal to each other:
sqrt( (a - 1)2 + (b)2 ) = sqrt( (a + 3)2 + b2 ) ---> (a - 1)2 + b2 = (a + 3)2 + b2
---> a2 - 2a + 1 + b2 = a2 + 6a + 9 + b2 ---> -8a = 8 ---> a = -1
Setting 1) and 3) equal to each other:
sqrt( (a - 1)2 + (b)2 ) = sqrt( (a)2 + (b - 1)2 } ---> (a - 1)2 + (b)2 = a2 + (b - 1)2
---> a2 - 2a + 1 + b2 = a2 + b2 - 2b + 1 ---> a = b ---> b = -1
Is there another answer?
+397 Questions like this can often be solved geometrically.
On the complex plane, | z - 1 | is the distance between the points z and (1, 0).
So, for example, if | z - 1 | = 1, the variable point z would be constrained to move on the circle centre (1, 0) with radius 1.
Here, | z - 1 | = | z + 3 |, says that z is equidistant from the points (1, 0) and (-3, 0) which means that it lies on the vertical line z = -1.
Similarly | z - 1 | = | z - i | says that z is equidistant from the points (1, 0) and (0, 1) meaning that it lies on the 45 deg line through the origin.
To satisfy both conditions, z must be at the intersection of these two lines, and that is easily seen to be the point (-1, -1).
