Verified answer

lz-1l=lz-il

z=a+bi

(a-1)^2+b^2=a^2+(b-1)^2

2a-1=2b-1

a=b

z=a+ai

so 1/z-1=1/(a-1)+ai

=(a-1)-ai

This is quite a readable introduction to complex loci

http://www.coventry.ac.uk/ec/jtm/6/dg6p6.pdf

Since lz-1l = lz-il

z lies on perpendicular bisector of line joining points 1 and i

Has general form z = k(1 + i) and slope 1

z - 1 = (k - 1) + ki

In general 1/(a + ib) = (a - bi)/(a^2 + b^2)

so 1/(z-1) = [(k - 1) - ki])/(2k^2 - 2k + 1)

This is a line with slope -k/(k - 1)

Regards - Ian

lz-1l=lz-il

z=a+bi

(a-1)^2+b^2=a^2+(b-1)^2

2a-1=2b-1

a=b

z=a+ai

so 1/z-1=1/(a-1)+ai

=[(a-1)-ai]/[(a-1)^2+a^2]