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Let $f(z) = \frac{2z}{z+1}$ and let $L_{1}$ be the $x$ axis.

I am told to show that $f(z)$ maps $L_{1}$ to itself.

I have picked three points to determine it image, $0,\ 1$ and $\infty$.

I'm having trouble as if $\infty$ gets mapped to $2$, how is the line $L_1$ mapped to itself? As the resulting mapping would give $0,1$ and $2$? And this is not $L_{1}$ as there is no point of $\infty$ to produce the line?

This is related to this question.

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harry55
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    In order to for this statement to be true, one must interpret the $x$-axis as the set of real-valued $z$ plus the point at infinity. Otherwise the mapping indeed does not send $L_1$ to itself. – Semiclassical Jan 05 '17 at 16:29

1 Answers1

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Mapping the $x$ axis to itself does not necessarily mean that any $x$ is mapped to the same $x$.

Our transformation maps $x$ to

$$\frac{2x}{x+1}.$$

The question is if this mapping is defined for all reals and if its range is the whole set of reals. The answer for the first question is no. Our function is not defined for $x=-1$.

As far as the range: If $y$, a member of the range, is given then for the corresponding $x$ we have

$$x=\frac y{2-y}.$$

So, $y=2$ is not in the range.

What we can say is that the transformation in question maps $\mathbb R\ \setminus \{-1\}$ to $\mathbb R\ \setminus\{2\}$.

zoli
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