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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?

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  • $\begingroup$ Why don't you tell us what result you got from transforming the imaginary axis, not just what the result isn't? $\endgroup$ – Rahul Feb 6 at 15:06
  • $\begingroup$ 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. $\endgroup$ – Kang Feb 6 at 17:40
  • $\begingroup$ 1 isn't on the imaginary axis as far as I know. $\endgroup$ – Rahul Feb 6 at 20:54
  • $\begingroup$ 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. $\endgroup$ – Kang Feb 7 at 7:14
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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.

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  • $\begingroup$ I used the inverse point equation of the imaginary axis as that was what Priestly using in this chapter. Your way is so much neater—and works! But I'd really like to know where I went wrong... $\endgroup$ – Kang Feb 6 at 17:43
  • $\begingroup$ I've just checked: the line you found is the same as mine—which has the inverse points 0 and 1+i. Problems arose when I tried out the transformation on some points on the imaginary axis which didn't seem to be on the line but I see now, thanks to your straightforward equation y=1-x, that they are. Thank you for clearing up my muddle! $\endgroup$ – Kang Feb 7 at 7:31

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