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Find the image of the $x$ and $y$ axes under $f(z)=\frac{z+1}{z-1}$

Atemppt

Notice that the real axis is given by $x+0i$ for $x\in \mathbb{R}$, then the image are $f(x)=\frac{x+1}{x-1}$ which is a hyperbola in the plane.

Now for the $y$ axis, we should have $0+iy$ but then $f(y)=\frac{iy+1}{iy-1}$ from here I can´t obtain a expression of in the plane, I tryed multiply $f$ by $\frac{iy-1}{iy-1}$ but it doesn´t work.

Too I tryed calculate $|f(y)|^2=1$, I think that it is a unit circle, but I don´t now if it least is sufficient. Any hint or comment was useful.

RobPratt
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Juan T
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  • $f$ maps the plane into the plane. Probably you are to describe the image of the horizontal axis in the plane. You seem to have jumped to a new $f$, that maps the real line into the real line. Your comment about hyperbola is useful, however. – 311411 Apr 16 '21 at 00:33
  • The book request me the following : At what angle do these images intersect?, and for that reason I think that the images are of the real and imaginary axis, for be specific I don´t know what exactly the book refer with the question What are the images of the $x$ and $y$ axes under $f$? – Juan T Apr 16 '21 at 00:44
  • I seen similar questions in stack, but in it posts only appear $f(z)=\frac{z-1}{z+1}$ – Juan T Apr 16 '21 at 00:45
  • For x-axis: your "notice that" above is right. Notice that if $z=x+0i$, then $f(z)$ is real. So in fact the correct answer for "the $f$-image of the x-axis" will be the range of real function $g(x)=\frac{x+1}{x-1}$, but viewed horizontally on the real axis. For the image of y-axis, you could apply theory of mobius transformation. (For example: https://math.stackexchange.com/questions/2084946) But I think you'll like Mr. Scott's answer below. – 311411 Apr 16 '21 at 00:52

2 Answers2

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If $x\in\Bbb R$, then $f(x)=\frac{x+1}{x-1}$ as a point in the complex plane, so the image under $f$ of the real axis is

$$\left\{\frac{x+1}{x-1}:x\in\Bbb R\right\}=\left\{1+\frac2{x-1}:x\in\Bbb R\right\}\,.$$

Plainly this is a subset of the real axis. Try to show that it is all of the real line except one point and identify the missing point.

For the image of the imaginary axis we want

$$\left\{\frac{yi+1}{yi-1}:y\in\Bbb R\right\}\,,$$

again as a subset of $\Bbb C$, the complex plane. Note that

$$\begin{align*} \frac{yi+1}{yi-1}&=\frac{yi+1}{yi-1}\cdot\frac{yi+1}{yi+1}\\ &=\frac{(yi+1)^2}{-y^2-1}\\ &=-\frac{1-y^2+2yi}{1+y^2}\\ &=\frac{y^2-1-2yi}{y^2+1}\,. \end{align*}$$

Now compute $\left|\frac{y^2-1-2yi}{y^2+1}\right|$ and verify that all of these points lie on the unit circle. Finally, show that they include all of the unit circle except one point, and identify the missing point.

Brian M. Scott
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$0\mapsto-1$, $1\mapsto \infty$, and $-1\mapsto0$ From that we know the image of the $x $-axis is itself.

$i\mapsto \dfrac {i+1}{i-1} $, and $-i\mapsto\dfrac {i-1}{1+i} $. Thus, since $0,i,-i $ map to three different points on the unit circle, the image of the $y $-axis is the unit circle.

In both cases the standard fact about Möbius transformations was used: namely they map generalized circles to generalized circles.