#1**+3 **

We are looking for all complex numbers that are equidistant from 3, -i and 3i. Since these three points are not collinear, we know there is a unique such point (the circumcenter of the triangle formed by the three numbers in the complex plane).

The numbers equidistant from and are on the perpendicular bisector of the segment connecting -i and 3i in the plane. This perpendicular bisector is Im(z)=1, so the imaginary part of is 1.

The numbers equidistant from 3 and 3i are on the perpendicular bisector of the segment connecting 3 and 3i in the plane. This perpendicular bisector is simply Re(z)=Im(z). Combining this with Im(z)=1 tells us that **z=1+i**.

Oofrence Oct 12, 2019

#2**+3 **

Find all complex numbers Z such that |z - 3| = |z+i| = |z -3i|

Let z=a+bi

So I want

\( |a+bi - 3| = |a+bi+i| = |a+bi -3i|\\ |(a-3)+bi | = |a+(b+1)i| = |a+(b-3)i |\\ (a-3)^2+b^2=a^2+(b+1)^2=a^2+(b-3)^2\\ a^2-6a+9+b^2=a^2+b^2+2b+1=a^2+b^2-6b+9\\ -6a+9=2b+1=-6b+9\\~\\ 2b+1=-6b+9\\ 8b=8\\ b=1\\~\\ -6a+9=2b+1\\ -6a+9=2*1+1\\ -6a+9=3\\ -6a=-6\\ a=1 \)

So the only z I get is \(\bf {z=1+i}\)

check

\( |z - 3| = |z+i| = |z -3i|\\ |1+i - 3| = |1+i+i| = |1+i -3i|\\ |-2+i | = |1+2i| = |1-2i|\\ \sqrt5 = \sqrt5 =\sqrt5\)

.Melody Oct 12, 2019