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# Complex Mappings watch

1. Can someone explain this to me?

I don't understand how get from say |z-1|<1 and then make the mapping z --> 1/z to get to an inverted circle (is that what it's meant to be?). I mean I can do it with |z|<1, that's easy it just becomes |z|>1 innit, but when I try and do the same with |z-1|<1 I get:

|1/z-1|<1

implies |1-z|<|z| implies |z-1|<|z| and this can't be an inverted circle because the point z=i does not satisfy the inequality and yet it's clearly outside the circle |z-1|<1 innit?

So what's hapenning?
2. (Original post by Cook)
Can someone explain this to me?

I don't understand how get from say |z-1|<1 and then make the mapping z --> 1/z to get to an inverted circle (is that what it's meant to be?). I mean I can do it with |z|<1, that's easy it just becomes |z|>1 innit, but when I try and do the same with |z-1|<1 I get:

|1/z-1|<1

implies |1-z|<|z| implies |z-1|<|z| and this can't be an inverted circle because the point z=i does not satisfy the inequality and yet it's clearly outside the circle |z-1|<1 innit?

So what's hapenning?
I would look at what happens to the boundary (ie, |z-1|=1). With these sorts of maps , circles or lines map to circles or lines. Just look at a few points to figure out the circle or line (in this case it's a line).
Then to check which part to shade in, just check some point inside the circle. eg, 1->1. Or alternatively, 1/z is never zero. (in this case, it's everything to the right of the line).
3. (Original post by SsEe)
I would look at what happens to the boundary (ie, |z-1|=1). With these sorts of maps , circles or lines map to circles or lines. Just look at a few points to figure out the circle or line (in this case it's a line).
Then to check which part to shade in, just check some point inside the circle. eg, 1->1. Or alternatively, 1/z is never zero. (in this case, it's everything to the right of the line).
But I don't understand how |z|<1 can be mapped to an inverted circle and |z-1|<1 can be mapped to a line under the SAME transformation, I mean they're both circles of radius one yet why are their images completely different?
4. Moebius maps take circles to circles.

(a line plus the point at infinity is defined to be a circle).
5. (Original post by Cook)
But I don't understand how |z|<1 can be mapped to an inverted circle and |z-1|<1 can be mapped to a line under the SAME transformation, I mean they're both circles of radius one yet why are their images completely different?
The mapping w --> 1/w isn't linear, or indeed particularly nice at all, so you shouldn't expect similar shapes to map to similar shapes. In particular, the boundary |z-1| = 1 contains the point 0, so under the mapping w --> 1/w, this will go away to infinity, and nearby points will stay 'nearby'. If that doesn't suggest automatically that you're going to get something like a line (or a parabola, or whatever - some curve that goes to infinity and doesn't quite join up with itself), I don't know what does!
6. (Original post by generalebriety)
The mapping w --> 1/w isn't linear, or indeed particularly nice at all, so you shouldn't expect similar shapes to map to similar shapes. In particular, the boundary |z-1| = 1 contains the point 0, so under the mapping w --> 1/w, this will go away to infinity, and nearby points will stay 'nearby'. If that doesn't suggest automatically that you're going to get something like a line (or a parabola, or whatever - some curve that goes to infinity and doesn't quite join up with itself), I don't know what does!
Ok that makes more sense now thanks. But what how would I go about doing the same thing for a more complicated mapping?

e.g. If I have the mapping z--> log(1+z) then what does |z|<1 get mapped to this time. Cos I can let z=exp([email protected]) say (cos that's roughly a complex circle innit?) and then

1+z=2exp([email protected]/2)cos(@/2)

and so using the definition of the complex logarithm:

log(1+z)=ln(2cos(@/2))[email protected]/2 but then what does this tell me??
7. (Original post by Cook)
Ok that makes more sense now thanks. But what how would I go about doing the same thing for a more complicated mapping?

e.g. If I have the mapping z--> log(1+z) then what does |z|<1 get mapped to this time. Cos I can let z=exp([email protected]) say (cos that's roughly a complex circle innit?) and then

1+z=2exp([email protected]/2)cos(@/2)

and so using the definition of the complex logarithm:

log(1+z)=ln(2cos(@/2))[email protected]/2 but then what does this tell me??
Well, not much. I'm sure you could probably find out the shape of that thing somehow if you really wanted. But why bother? You're clearly considering Möbius maps in the rest of your posts, which are a much nicer group of functions which always map circles (and lines) to circles (and lines). No need to deal with anything complicated, because Möbius maps aren't complicated.
8. (Original post by generalebriety)
Well, not much. I'm sure you could probably find out the shape of that thing somehow if you really wanted. But why bother? You're clearly considering Möbius maps in the rest of your posts, which are a much nicer group of functions which always map circles (and lines) to circles (and lines). No need to deal with anything complicated, because Möbius maps aren't complicated.
That's a very defeatist attitude generalebriety! I don't just want to consider mobius maps (nice as they are life just aint that simple), I want to consider conformal mappings in general.

Going back to my example of log(1+z) and the unit disk |z|<1, I know for exp([email protected]) the principal range for @ is -pi<@<pi and so do I just vary @ in the expression I got for log(1+z) and see what range of values it can take? But then how does the |z|<1 come into this? What if it was |z|>1?

And does the fact that log(1+z) has a branch point at z=-1 have anything to with anything? (If so, let's say for the sake of argument we take the branch cut on the real axis from -infinity to -1).
9. (Original post by Cook)
That's a very defeatist attitude generalebriety! I don't just want to consider mobius maps (nice as they are life just aint that simple), I want to consider conformal mappings in general.
It's not really a defeatist attitude; the fact is that you were getting confused about Möbius maps earlier. How about you learn to walk before you can run?

(Original post by Cook)
Going back to my example of log(1+z) and the unit disk |z|<1, I know for exp([email protected]) the principal range for @ is -pi<@<pi and so do I just vary @ in the expression I got for log(1+z) and see what range of values it can take?
Yes. How you do that is up to you, but yes, that is just a restatement of your problem.

(Original post by Cook)
But then how does the |z|<1 come into this? What if it was |z|>1?
As you were advised to do before, you'd do well to look at the image of the boundary |z| = 1 first before worrying about its interior. The image of the boundary should hopefully be the boundary of the image (if your map is nice enough), and then you just need to fiddle about to find out which bit is |z| < 1 and which is |z| > 1. I'm being deliberately vague here, partly because I can't be bothered to work out the answer myself because I don't think it's very interesting, partly because I think you should learn about Möbius maps before learning about the general case (a full description of which probably doesn't exist).

(Original post by Cook)
And does the fact that log(1+z) has a branch point at z=-1 have anything to with anything? (If so, let's say for the sake of argument we take the branch cut on the real axis from -infinity to -1).
I don't see why it would. More important is the fact that z = -1 is a pole of that function. But my complex analysis is rusty (to be generous to it).
10. (Original post by generalebriety)
It's not really a defeatist attitude; the fact is that you were getting confused about Möbius maps earlier. How about you learn to walk before you can run?

Yes. How you do that is up to you, but yes, that is just a restatement of your problem.

As you were advised to do before, you'd do well to look at the image of the boundary |z| = 1 first before worrying about its interior. The image of the boundary should hopefully be the boundary of the image (if your map is nice enough), and then you just need to fiddle about to find out which bit is |z| < 1 and which is |z| > 1. I'm being deliberately vague here, partly because I can't be bothered to work out the answer myself because I don't think it's very interesting, partly because I think you should learn about Möbius maps before learning about the general case (a full description of which probably doesn't exist).

I don't see why it would. More important is the fact that z = -1 is a pole of that function. But my complex analysis is rusty (to be generous to it).
I am not just considering mobius maps I am actually considering conformal maps in general, specifically logz. But it's been a while since I've done them and I was just having a very annoying mind block so I needed someone to sort out that first before we came on to more compicated stuff.

i.e. if you could just explain to me what happens to say the circle |z|<2 under the map z-->logz that would be very helpful. If (like you say) we consider the boundary z=2exp([email protected]) then log(2exp([email protected])=ln(2) [email protected], does this mean that the boundary of the circle gets mapped to the finite line where the real part is ln2 and the imaginary part can take values in -pi<@<pi? And what about the rest of the circle where does that go?
11. (Original post by Cook)
I am not just considering mobius maps I am actually considering conformal maps in general, specifically logz. But it's been a while since I've done them and I was just having a very annoying mind block so I needed someone to sort out that first before we came on to more compicated stuff.

i.e. if you could just explain to me what happens to say the circle |z|<2 under the map z-->logz that would be very helpful. If (like you say) we consider the boundary z=2exp([email protected]) then log(2exp([email protected])=ln(2) [email protected], does this mean that the boundary of the circle gets mapped to the finite line where the real part is ln2 and the imaginary part can take values in -pi<@<pi? And what about the rest of the circle where does that go?
As I say, my complex analysis is rusty. However, remember that log is a multifunction - that is, even though z = 2exp(it) (pi < t <= pi) is a closed curve, log has a branch point at 0, and so either its image under log won't be closed or you're expected to do some trickery by letting t vary over all of R - I don't really know.

(Imagine an infinite spiral S, similar to this one, with an obvious projection onto the complex plane C (minus the point at 0). Then log can be made a well-defined function on S, but obviously when we take the projection of S onto C, things go wrong, because going one way round the origin corresponds to going up the spiral, and going the other way corresponds to going down. Very similar to taking a branch cut. Surfaces like this are called Riemann surfaces.)

Anyway, once you know the answer to this, you know where the rest of the circle goes, because the rest of the circle is just z = s*exp(it) for 0 <= s < 2.

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