Mobius transformation produces either a circle or a line...

Question

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?

Answer 1

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

One can see that it is an equation of a line.