I know that the mapping -1/z is conformal away from the origin, since the mapping would then be analytic and have a non-zero derivative everywhere in C.

It apparently also maps the upper half plane conformally onto itself - but how can I see this?

I used a few test points and saw that 0 maps to infinity, infinity maps to 0, 1 maps to -1, i maps to i.

Then I actually use a continuity argument - and not anything from complex analysis - to justify that this mapping must map the upper half plane onto itself: the upper half plane is a connected set, and the mapping -1/z is continuous away from the real axis, which contains the problem point, z=0, and continuous functions must map connected sets to connected sets. So, we conclude that -1/z must map the upper half plane onto itself.

Is there a better / correct / complex analysis way to learn what certain mappings do? As I wrote above, I noticed that the point i maps to i ... could I translate this into a complex analysis argument that -1/z must map the upper half plane conformally onto itself?

I have heard from a student that "any mapping that fixes the real line also fixes the respective planes". Is this perhaps another way of thinking about the action of conformal mappings? So, by his statement (wherever he found this theorem from), then mappings such as 1/z, -1/z, 1/z^2, 1/z^3, and other mappings that clearly fix the real axis ...are conformal mappings of the upper half plane to itself -- and conformal mappings of the lower half plane to itself.

Any help would be greatly appreciated.

Thanks in advance,

Edit 4: Another thought about the mapping -1/z is: I'm struggling to know for certain that it is a conformal mapping. It is certainly pointwise conformal away from the real axis, but to be a conformal mapping, it needs to be holomorphic, have non-zero derivative, as well as be a one-to-one and onto mapping. I don't know for certain that -1/z is bijective from the upper half plane to itself...

Edit 3: Something is very wrong with the 'theorem' that I cited above, because 1/z maps i to -i ... so this mapping doesn't fix the upper and lower half planes, even though it fixes the real axis.

Edit 2: Actually, I wouldn't know the image, even though I know it's a connected set. So, I guess the test point z=i mapping to itself was crucial.

Edit: To go a bit further with that student's statement, then any mapping that fixes the imaginary axis fixes the left half plane and right half plane.


Let $z=x+iy$ such that $y>0$. Then, $z^{-1}= \frac{1}{z} = \frac{1}{|z|^2}\bar{z} = \frac{1}{x^2+y^2}x-iy$. This implies that: $$ -\frac{1}{z} = -\frac{1}{x^2+y^2}(x-iy) = \frac{1}{x^2+y^2}(-x+iy). $$ Since $y>0$, this number also lies in the upper plane.

  • $\begingroup$ Very nice argument, @hjhjhj57 :) Do you think I should just have a bunch of maps memorized? I have a few complex analysis books and Brown and Churchill's book has a table of mappings in the back of it. Or, should I continue to learn about conformal mappings from more an intuition angle and justify the mappings (like what you just showed) ...? I'm pretty new to this topic, so I'm not sure what is the best way to proceed with practicing for these types of problems that use conformal mappings... $\endgroup$ – User001 Dec 27 '14 at 7:36
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    $\begingroup$ @LebronJames Of course it depends on what you want, but I'd focus on learning the basic theory. Any decent book on Complex Analysis should have a section on conformal mappings with all the rudiments, e.g. Möbius transformations (of which your question is just an example) or mentioning that every biholomorphism between two regions is conformal. $\endgroup$ – hjhjhj57 Dec 27 '14 at 7:42
  • $\begingroup$ Ok, got it. Thanks so much, @hjhjhj57. Have a great night :) $\endgroup$ – User001 Dec 27 '14 at 7:46
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    $\begingroup$ @LebronJames Have you tried using the same argument? What does your function do to $(-\infty,0)$ and $(0,\infty)$? $\endgroup$ – hjhjhj57 Dec 27 '14 at 9:03
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    $\begingroup$ @LebronJames you're right. But formally you can't say it fixes the real line, since it isn't defined in $0$. It fixes the positive and negative real axes. On the other hand, if you're working on the extended complex plane, it does send the real line to itself. $\endgroup$ – hjhjhj57 Dec 28 '14 at 1:26

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