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This might be an odd question, but I recently spent a very long time solving a problem in Gamelin's Complex Analysis and I wanted to understand exactly what was making it so difficult for me. Eventually I realized it was that I didn't quite understand this passage from Chapter II.7

Consider first a circle that does not pass through $\infty$. It has an equation of the form $|z-a|^2 = r^2$, where $a$ is its center and $r$ is its radius. The image of the circle under the inversion $w = 1/z$ consists of points satisfying $|1-aw|^2 = r^2|w|^2$.

Now this all seems simple enough, but what I realized was that I wasn't sure you could just do what Gamelin does above.

I.e. if you have an equation/variety $f(z) = 0$, and then you do some transformation $z \mapsto w = g(z)$, is it always the case that $f(w) = f(g(z)) = 0$? Obviously this is true for when $g$ is a coordinate transformation, but when isn't it true? I guess I should really be asking: When is a function a coordinate transformation and when is it not?

I apologize if this is too vague or obvious, it just seems like it shouldn't be the case that you can always do this, like there must be some restrictions on $f$ and $g$.

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In this case $g: \mathbb{C}-0 \to \mathbb{C}-0$ is a bijection and $g\circ g = id$. So the image of $$\{z \in \mathbb{C}-0: f(z)=0\}$$ under $g$ is $$\{w \in \mathbb{C}-0: f(g^{-1}(w))=0\} =\{w \in \mathbb{C}-0: f(g(w))=0 \}$$ since $g$ is equal to its inverse.

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  • $\begingroup$ Very nice, thank you! $\endgroup$ Commented Jun 16, 2018 at 3:14
  • $\begingroup$ For $x \in \mathbb{C}$, let $f(z) = z-x$. Then for $g: \mathbb{C} \to \mathbb{C}$, $\{z: f(g(z)) = 0\}$ equals $\{x\}$ implies $g(g(x)) = x$. Hence for fixed $g$, the condition holds for all $f$ if and only if $g$ is its own inverse. $\endgroup$ Commented Jun 16, 2018 at 3:17
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You've actually got it backwards: if $g:A \to B$ and $f: A \to C$, the only composition that could make sense in general is $f \circ g^{-1}$. In this case, $g^{-1}=g$, which stopped you from spotting the general law. The solutions are the set $\{ w: \exists z : g(z)=w \wedge f(z) = a \}$, so if these can be solved simultaneously, you can find an explicit equation.

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