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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.

 Apr 9, 2022
 #1
avatar+23183 
+2

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?  

 Apr 9, 2022
 #2
avatar+225 
+1

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).

 Apr 9, 2022

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