# Mobius transformation produces either a circle or a line…

The exercise (from H A Priestley) required a transformation that sent $$\:0, 1, {\infty}$$ to $$1, 1+i, i$$. I knew the transformation that sent $$z_1, z_2,z_3,$$ to $$0, 1, {\infty}$$ ie $$\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$$ So I found the inverse using $$z_1=1,\ z_2=1+i,\; z_3=i\;$$ which I made $$\frac{(z+1)}{(1-iz)}$$ The question required one to use this transformation on various objects which all seemed to work perfectly until I came to the last one which was the imaginary axis. The result seemed to be neither a circle (it went through $${\infty}$$) nor a straight line. Brains have been racked in vain: please, where have I gone wrong?

• Why don't you tell us what result you got from transforming the imaginary axis, not just what the result isn't? – Rahul Feb 6 at 15:06
• Sorry. I got infinity for -i, 1 for zero and 1+i for 1. Which would be fine. But for ki where k is bigger than 1 or less than -1 the points aren't on that line at all. – Kang Feb 6 at 17:40
• 1 isn't on the imaginary axis as far as I know. – Rahul Feb 6 at 20:54
• Sorry—muddle: I should have said that 1 and -1, the inverse points for the imaginary axis, went to 0 and i+i, giving a line that is the perpendicular bisector of 0 to 1+i. Then I thought I'd try a few imaginary points with the transformation—and found 2i, 3i etc weren't on the line. – Kang Feb 7 at 7:14

Imaginary axis has the equation $$z=it$$. After transformation new curve will have the equation $$z=\frac{1+it}{1+t},\qquad x=\mathrm{Re}\ z =\frac{1}{1+t},\qquad y=\mathrm{Im}\ z =\frac{t}{1+t} = 1-x$$