SOLUTION: 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.

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: 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.      Log On


   

Question 1074905: 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.
Found 2 solutions by Fombitz, ikleyn:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
z=a%2Bbi
So,
abs%28z-1%29=sqrt%28%28a-1%29%5E2%2Bb%5E2%29
abs%28z-3%29=sqrt%28%28a-3%29%5E2%2Bb%5E2%29
abs%28z-i%29=sqrt%28a%5E2%2B%28b-1%29%5E2%29
.
.
%28a-1%29%5E2%2Bb%5E2=%28a-3%29%5E2%2Bb%5E2
a%5E2-2a%2B1%2Bb%5E2=a%5E2-6a%2B9%2Bb%5E2
-2a%2B1=-6a%2B9
4a=8
a=2
and
%28a-1%29%5E2%2Bb%5E2=a%5E2%2B%28b-1%29%5E2
a%5E2-2a%2B1%2Bb%5E2=a%5E2%2Bb%5E2-2b%2B1
-2a%2B1=-2b%2B1
b=a
b=2
.
.
.
z=2%2B2i

Answer by ikleyn(52772) About Me  (Show Source):
You can put this solution on YOUR website!
.
The problem asks for a point z which in the complex plane is equidistant from complex points A = (1,0), B = (-3,0) and C = (0,i).


This point is actually the center of the circumscribed circle about these three points.


I can find this point without calculations (mentally).


The center of the circumscribed circle lies at the intersection of perpendicular bisectors to the sides. 


The perpendicular bisector to the side AB is x = -1.


The perpendicular bisector to the side AC is the straight line y = x.


Their intersection is the point (-1,-1), which is the complex number -1-i.


Answer.  The complex number under the question is -1-i.