A set in the complex plane edit: ?s about complex variables

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A set in the complex plane edit: ?s about complex variables

Postby agelessdrifter » Wed Jun 08, 2011 12:01 am UTC

This question comes from a "homework" question in an intro to Complex Analysis text, but it's not homework, I'm just studying independently.

The question is meant to test whether I understand the definitions of open, closed, bounded and connected by providing example sets and asking me to identify which properties the sets satisfy. One example set was the following:

[imath]\{ z \in C : |z+1| + |z-1| = 2\}[/imath]

Now I couldn't figure out what z, in general, would satisfy this condition, and I couldn't think of a way to simplify it algebraically to make it obvious, so I just tried to find a few example points that I knew would satisfy it to see if I could extrapolate a pattern from them. I decided that [imath]\{z|z=x+iy,-1\leq x\leq 1, y=0\}[/imath] satisfied the condition, but I didn't think that could be the whole solution, so I tried wolfram alpha, which seems to verify that it is.

My question is whether there is some algebraic or even a general intuitive way to reach this conclusion. I feel like there can't be, since the solution includes the requirement that y=0, and by using the definition of the modulus, I eliminate any i's from the expression, which makes it seem unlikely that I'd be able to show that the imaginary component must be zero.

(Couldn't figure out how to input the blackboard C for the set of complex numbers -- I tried \mathbb{C} but it wasn't recognized)
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Re: A set in the complex plane

Postby The Sleeping Tyrant » Wed Jun 08, 2011 12:28 am UTC

The way I would look at this is by fixing some r <= 2, and then look at the points with [imath]|z + 1| = r[/imath] and [imath]|z - 1| = 2 - r[/imath]. So we get some points in your set by looking at the intersection of these two circles, which just happens to be the point (r - 1, 0). So, look at each r between 0 and 2, and your set will be [imath]\{(r - 1,0)| 0 \le r \le 2\}[/imath], which is the set you gave.

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Re: A set in the complex plane

Postby MostlyHarmless » Wed Jun 08, 2011 12:34 am UTC

I would think of it like this: Your set is the set of all points such that the distance from -1 and the distance from 1 add up to 2. That is the definition of an ellipse with foci at -1 and 1, so you would normally expect an elliptical shape. (To see this, imagine taking a string 2 units long and fixing the ends at your foci. Hold onto the string and pull it out taught. As you move your finger, you will trace out an ellipse.) However, this is a degenerate ellipse, since the foci are exactly 2 units apart, so the string is already taught. Therefore, you only get points along the line between your two foci.

Edit: Here's a video to explain what I mean. http://www.youtube.com/watch?v=7UD8hOs- ... re=related
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Re: A set in the complex plane

Postby skeptical scientist » Wed Jun 08, 2011 12:38 am UTC

{ z in C : |z+1| + |z-1| = 2} is the set of points z such that the distance from z to 1, plus the distance from z to -1, equals 2. Since the shortest distance between two points is a straight line, the distance from 1 to -1 through z is at least 2, and is exactly 2 iff z is on the line segment between -1 and 1.
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Re: A set in the complex plane

Postby agelessdrifter » Wed Jun 08, 2011 12:43 am UTC

Took me a second to see your reasoning, but that's an interesting approach. I guess I'll practice that with some similar sets to make sure I can apply it.


edit: Hmm. The solution was equally simple when I tried |z+1|+|z+3|=2, but became much more complicated when I tried |z+2|+|z-i|=3. I don't think I would have figured out a general approach like that on my own.

2nd edit: Oh, I'm sorry, I made this post originally after reading only The Sleeping Tyrant's response. Thanks for the subsequent explanations.
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Re: A set in the complex plane

Postby Tirian » Wed Jun 08, 2011 1:21 am UTC

MostlyHarmless wrote:taught


"taut"

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Re: A set in the complex plane

Postby MostlyHarmless » Wed Jun 08, 2011 7:33 am UTC

D'oh. I feel stupid.

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Re: A set in the complex plane

Postby agelessdrifter » Sun Jun 12, 2011 11:10 pm UTC

I edited the title of the thread because I think I am going to continue to have quick questions during this course, and don't want to bombard the forum with them, thread-by-thread.

Currently I am working on an exercise identifying the regions of differentiability and of "analyticity" of given functions. My weapon of choice for this exercise seems like it ought to be the following theorem:

(a) Suppose [imath]f[/imath] is di fferentiable at[imath]z_0 = x_0 + iy_0[/imath]. Then the partial derivatives of [imath]f[/imath] satisfy [imath]\frac{\partial f}{\partial x}= -i\frac{\partial f}{\partial y} (2.1)[/imath].

(b) Suppose[imath]f[/imath] is a complex function such that the partial derivatives [imath]f_x[/imath] and [imath]f_y[/imath] exist in an open
disk centered at [imath]z_0[/imath] and are continuous at [imath]z_0[/imath]. If these partial derivatives satisfy (2.1) then f is differentiable at [imath]z_0[/imath].

In both cases (a) and (b), [imath]f'[/imath] is given by [imath]\frac{\partial f}{\partial x}(z_0)[/imath]


Which is to say my approach has been to find [imath]\frac{\partial f}{\partial x}[/imath] multiply it by i, and set that equal to [imath]\frac{\partial f}{\partial y}[/imath], then simplify the expression to identify in what region this relationship can be true, since if it is not true in a region, then f must not be differentiable in that region. Then, by identifying the region in which these first partials are continuous, and taking the intersection with the above-described region, I find my region of differentiability. Maybe this is a bad approach -- either way I am obviously misapplying something. Here is an example:

[imath]f(z)=z\Im m (z) = xy+iy^2[/imath] The given solution is that [imath]f[/imath] is differentiable at z=i with derivative i, and nowhere analytic.

My approach was as follows:

[math]if_x(z)=f_y(z)
\Rightarrow
i\frac{\partial}{\partial x}(xy+iy^2)=\frac{\partial}{\partial y}(xy+iy^2)[/math]
[math]\Rightarrow iy = x+2iy \Rightarrow y=ix[/math]

And since [imath]x,y \in \Re[/imath], this can't be true for any value of x and y other than x=y=0 (and [imath]f_y[/imath] and [imath]f_x[/imath] are continuous at the origin), therefore [imath]f[/imath] is differentiable at the origin.

z=i would mean 1=0i, so either the given answer is wrong, or my approach is way off (more likely).

Can someone tell me why this isn't working? Thanks in advance. I'm starting to become nervous about pursuing a degree in math.


edit: Additionally, apropos to the above example, whether my reasoning for the region of differentiability is correct or not, the theorem plainly states that the derivative of f is equal to the partial derivative of f with respect to x, which in this case would simply be y. Since there's no 'i' there, the derivative can't possibly equal i. So maybe the given solution is wrong after all?

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Re: A set in the complex plane edit: ?s about complex variab

Postby jestingrabbit » Sun Jun 12, 2011 11:59 pm UTC

I went back to first principles and got the same answer as you. Furthermore, [math]\frac{f(i+he^{i\theta} ) - f(i)}{he^{i\theta}} = 1 + h\sin \theta +i\frac{\sin\theta}{e^{i\theta}}[/math] and you can see that the limit as h goes to 0 will vary depending on what theta is.

You were right, the book was wrong, the Cauchy Riemann equations are the right thing to apply.
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Re: A set in the complex plane edit: ?s about complex variab

Postby agelessdrifter » Mon Jun 13, 2011 12:27 am UTC

Thanks, that is a relief. Several of the answers in this text have simply not made any sense and I was really starting to worry. It's an e-book that my professor gave to me -- maybe I got a poorly-edited early version or something.

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Re: A set in the complex plane edit: ?s about complex variab

Postby jestingrabbit » Mon Jun 13, 2011 5:40 am UTC

agelessdrifter wrote:Thanks, that is a relief. Several of the answers in this text have simply not made any sense and I was really starting to worry. It's an e-book that my professor gave to me -- maybe I got a poorly-edited early version or something.


That's possible. The text I have for this stuff is Churchill and Brown, and its perfectly adequate, though it had a non zero cost as opposed to the text you have. You might be able to get a newer edition of what you've got somewhere.
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Re: A set in the complex plane edit: ?s about complex variab

Postby agelessdrifter » Tue Jun 14, 2011 6:12 pm UTC

I asked my professor this question today, but his answer was sort of hurried and ambiguous, and I'm not sure if he understood me correctly:

Since there are limitations on the types of functions for which we can find (df/dz) (where z=x+iy) in the complex plane, as a result of the need to prove that the limit of the difference quotient is equivalent along any approach, does it make sense/is it useful to talk about the directional derivative of a function of z (thinking particularly of cases wherein df/dz can't be found)? That way we would only need to prove the limit to be equal from two directions, and we would at least know something about the rate of change of z at a point instead of nothing.

What I asked my professor was, roughly "Does it make sense to talk about directional derivatives in the complex plane?", and he responded "not really, no", but I'm not sure if he thought I meant "does it make sense for for us to talk about directional derivatives today/in this class", since at the time he was sort of perusing the literature and deciding what to lecture on for the day since I was the only one in class.

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Re: A set in the complex plane edit: ?s about complex variab

Postby ++$_ » Tue Jun 14, 2011 8:46 pm UTC

You can talk about directional derivatives in the complex plane, but it does not yield any interesting results as far as I know. The functions of interest are the analytic ones.
agelessdrifter wrote:That way we would only need to prove the limit to be equal from two directions
Whether you are talking about real numbers or complex numbers, you need to consider all paths, not just two (and in fact, you can even have a function where all directional derivatives are defined at a point, but the function is not differentiable at that point).

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Re: A set in the complex plane edit: ?s about complex variab

Postby agelessdrifter » Tue Jun 14, 2011 10:08 pm UTC

I meant that for the directional derivative, you only need the limit of the difference quotient from two directions. Is that not so?

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Re: A set in the complex plane edit: ?s about complex variab

Postby ++$_ » Tue Jun 14, 2011 10:56 pm UTC

Only if the function is differentiable at that point.

To find out if the function is differentiable, you have to do the work you were trying to avoid anyway.

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Re: A set in the complex plane edit: ?s about complex variab

Postby Dopefish » Tue Jun 14, 2011 11:16 pm UTC

you can even have a function where all directional derivatives are defined at a point, but the function is not differentiable at that point


Example and/or relevant link plox? My intuition doesn't like that statement, but then I'm not a mathematician and you folks can come up with some really weird functions.

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Re: A set in the complex plane edit: ?s about complex variab

Postby agelessdrifter » Wed Jun 15, 2011 12:42 am UTC

I guess I'm fuzzy on the idea of directional derivatives (it has been a while since I've taken multivariable calc). I feel like this

you can even have a function where all directional derivatives are defined at a point, but the function is not differentiable at that point



contradicts this.
++$_ wrote:Only if the function is differentiable at that point.


I mean, the directional derivative (thinking of the three-dimensional case here) gives the rate of change of z (where z is just a real number in this case) as we move an infinitesimal distance from a point of origin along a path parallel to some vector, right? Why would we care about the limit approaching from directions other than the one along which that vector points?

I (always) feel like maybe I'm communicating poorly, so to illustrate what I mean here,

Image

If we're interested in the directional derivative of a function at [imath](x_0,y_0)[/imath] for the direction defined by U, why do we care about the limit approaching on path W? We're only interested in the infinitesimal change in z given an infinitesimal move along U, right? It makes sense to me that we'd care about the approach along U and along -U -- those are the two directions I had in mind.

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Re: A set in the complex plane edit: ?s about complex variab

Postby ++$_ » Wed Jun 15, 2011 3:09 am UTC

Dopefish wrote:
you can even have a function where all directional derivatives are defined at a point, but the function is not differentiable at that point
Example and/or relevant link plox? My intuition doesn't like that statement, but then I'm not a mathematician and you folks can come up with some really weird functions.
Consider the function in the real plane x3y/(x6 + y2) (and define it to be 0 at 0). Plug in your favorite line through (0,0) and you will see that the derivative along that line is defined. However, the function is not even continuous at the origin (and therefore not differentiable at the origin) -- put y = x3 to see this.
agelessdrifter wrote:It makes sense to me that we'd care about the approach along U and along -U -- those are the two directions I had in mind.
Aha. Yes, if we are only interested in the function in some particular direction, the calculus gets a lot easier.

However, it's not a very interesting thing to be interested in. The complex derivative has natural interpretations and properties (if you're taking complex analysis, you will meet some of these shortly), whereas the directional derivative doesn't really.

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Re: A set in the complex plane edit: ?s about complex variab

Postby agelessdrifter » Thu Jun 30, 2011 10:33 pm UTC

Ok, here are some problems I am looking at at the moment that are sort of stumping me:
(I typed out a big long post in latex an hour or two ago but lost it because I had to step away for a minute, so I'm just gonna screen-cap the questions:
Image

Question 13 in Chapter 2 was a proof that a function which is analytic and always real valued in G must be constant in G.

Firstly, in question 6, I'm interpreting "[imath]f(G)[/imath] is a subset of the unit circle" to mean [imath]\{f(z)|z \in G \} \subset \{z: |z|=1 \}[/imath], and I'm not sure if that's correct or not (to be clear, I dunno exactly what it means to take the function of a region -- I've never encountered that before in that particular representation).

But assuming that's correct, I can see that, since I proved that [imath]w(z)=\frac{1+z}{1-z}[/imath] maps the unit circle to the imaginary axis, if f(G) is a subset of the unit circle, then [imath]w(f(G))=\frac{1+f(G)}{1-f(G)}[/imath] is going to give me a subset of the imaginary axis, and from there, it's simple to prove that the function must be constant by using the Cauchy-Reimann equations. But that just tells me about w being constant, doesn't it? I don't see how that's useful in saying anything about f.

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Re: A set in the complex plane edit: ?s about complex variab

Postby Lothar » Fri Jul 01, 2011 12:02 am UTC

agelessdrifter wrote:Firstly, in question 6, I'm interpreting "[imath]f(G)[/imath] is a subset of the unit circle" to mean [imath]\{f(z)|z \in G \} \subset \{z: |z|=1 \}[/imath], and I'm not sure if that's correct or not (to be clear, I dunno exactly what it means to take the function of a region -- I've never encountered that before in that particular representation).

That would be my interpretation as well. It's quite common to write f(G) for the image of G under f. Similarly, f-1(Y) is the set of values mapped to Y by f.

agelessdrifter wrote:But assuming that's correct, I can see that, since I proved that [imath]w(z)=\frac{1+z}{1-z}[/imath] maps the unit circle to the imaginary axis, if f(G) is a subset of the unit circle, then [imath]w(f(G))=\frac{1+f(G)}{1-f(G)}[/imath] is going to give me a subset of the imaginary axis, and from there, it's simple to prove that the function must be constant by using the Cauchy-Reimann equations. But that just tells me about w being constant, doesn't it? I don't see how that's useful in saying anything about f.

If [imath]\frac{1+f(z)}{1-f(z)} = c[/imath], can you find an expression for f(z) in terms of c?
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Re: A set in the complex plane edit: ?s about complex variab

Postby agelessdrifter » Mon Jul 04, 2011 2:40 am UTC

Thanks, Lothar. That should have been really obvious :|

Right now I'm working on cross-ratios. The last problem was to take a given circle, come up with three different mobius transformations that map it to the real axis + infinity, and calculate the image of the center of the given circle under each transformation. I guess I was supposed to detect a pattern as to how the selected values of [imath]z_k[/imath] affected the outcome of the transformation of the circle's center, because now I am supposed to come up with a transform that maps a given circle onto itself, but maps that circle's center onto x=1/2, and I haven't a clue how to do that. If I've been reading this chapter right, I need to use three points on the circle to make sure the transformation produces the same circle -- if I use two points from the circle + the circle's center to define the function I'll wind up with completely different circle, I think.

I mean I guess I could come up with a system of equations in the fashion of [imath]f(z)=\frac{az+b}{cz+d}, f(0)=\frac{1}{2}, f(i)=1, f(1)=-i, f(-i)=-1[/imath] and solve for a,b,c and d, but I'm certain there's an easier method I'm supposed to apply here.

edit: I think I might have that one figured out.

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Re: A set in the complex plane edit: ?s about complex variab

Postby jestingrabbit » Mon Jul 04, 2011 12:17 pm UTC

agelessdrifter wrote:Thanks, Lothar. That should have been really obvious :|

Right now I'm working on cross-ratios. The last problem was to take a given circle, come up with three different mobius transformations that map it to the real axis + infinity, and calculate the image of the center of the given circle under each transformation. I guess I was supposed to detect a pattern as to how the selected values of [imath]z_k[/imath] affected the outcome of the transformation of the circle's center, because now I am supposed to come up with a transform that maps a given circle onto itself, but maps that circle's center onto x=1/2, and I haven't a clue how to do that. If I've been reading this chapter right, I need to use three points on the circle to make sure the transformation produces the same circle -- if I use two points from the circle + the circle's center to define the function I'll wind up with completely different circle, I think.

I mean I guess I could come up with a system of equations in the fashion of [imath]f(z)=\frac{az+b}{cz+d}, f(0)=\frac{1}{2}, f(i)=1, f(1)=-i, f(-i)=-1[/imath] and solve for a,b,c and d, but I'm certain there's an easier method I'm supposed to apply here.

edit: I think I might have that one figured out.


Assuming that the circle is the unit circle centred at the origin, the way that I would do this question is pick some transform that maps the circle to the line, say f, and then work out what f(0) and f(1/2) are. You then need a transformation, g, that fixes the line and is such that g(f(0)) = f(1/2) (hint: you can pick all the constants that define g so that they are in a simple set, and you can make an even stricter restriction on g). Then the map that you want is [imath]f^{-1} \circ g \circ f[/imath].
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Re: A set in the complex plane edit: ?s about complex variab

Postby gorcee » Wed Jul 06, 2011 3:32 am UTC

jestingrabbit wrote:
agelessdrifter wrote:Thanks, that is a relief. Several of the answers in this text have simply not made any sense and I was really starting to worry. It's an e-book that my professor gave to me -- maybe I got a poorly-edited early version or something.


That's possible. The text I have for this stuff is Churchill and Brown, and its perfectly adequate, though it had a non zero cost as opposed to the text you have. You might be able to get a newer edition of what you've got somewhere.


Describing C&B's cost as "non-zero" is like, the understatement of the year.

Great book, but one of the most pricey per page that I ever bought.

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Re: A set in the complex plane edit: ?s about complex variab

Postby agelessdrifter » Sat Jul 16, 2011 3:04 am UTC

jestingrabbit wrote:Assuming that the circle is the unit circle centred at the origin, the way that I would do this question is pick some transform that maps the circle to the line, say f, and then work out what f(0) and f(1/2) are. You then need a transformation, g, that fixes the line and is such that g(f(0)) = f(1/2) (hint: you can pick all the constants that define g so that they are in a simple set, and you can make an even stricter restriction on g). Then the map that you want is [imath]f^{-1} \circ g \circ f[/imath].


Thanks, that wound up working.

Right now I'm reading about homotopy and I have a quick question (actually I'm a ways past it now, but this question has been bugging me). This is the definition of a homotopy in the text (again I took a snapshot because it's easier than retyping it up in LateX -- I hope that's not problematic)

Image

My question is basically, why all the zeros and ones? Why not 'a's and 'b's? What if you have two curves parametrized by [imath]\gamma_1(t)[/imath] with t between a and b, and a continuous function [imath]h: [a,b]^2\rightarrow G[/imath] such that h(t,a)=[imath]\gamma_1(t)[/imath] and h(t,b)=[imath]\gamma_2(t)[/imath] and for all s, h(a,s)=h(b,s)? It just seems out of character for them not to be that general if it's possible to do so without something in the mechanics breaking down as a result, but I can't see why that wouldn't work just as well.

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Re: A set in the complex plane edit: ?s about complex variab

Postby mfb » Sat Jul 16, 2011 10:47 am UTC

It does not matter whether you use a<=t<=b or 0<=t<=1 as parameter for your curves.
Note that the theorem tells you something about the curves itself, not about some parameterizations of them.

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Re: A set in the complex plane edit: ?s about complex variab

Postby Yakk » Sat Jul 16, 2011 11:52 am UTC

They use zeros and ones because using [a,b] doesn't add anything interesting, while it just adds complexity.

Really, I don't think they don't even need the full power of the real continuum to say the things they want to say, but rather than describing the "least continuum" (ie, most general) that they want, they just use a fixed interval [0,1], because the details of the "set being mapped from" isn't the interesting part they want to work with.
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Re: A set in the complex plane edit: ?s about complex variab

Postby agelessdrifter » Tue Jul 19, 2011 1:30 am UTC

Hmm I had a feeling that was all it came down to, but I wanted to be sure. It wasn't the parameter t being arbitrarily bounded that concerned me so much as the parameter s. Anyway it still seems a little strange that they'd do that, but I trust you all.

Today we looked at the integral

[math]\int_0^\infty{\frac{1}{x^4+1}dx}[/math]

Which was a bit confusing without having gotten to the proof of the residues theorem in the text, but I was able to make of sense of it taking the RT on faith for the time being. The only thing I'm a bit confused about is why we only used the residues in the upper half of the plane.Is it just because the curve we created (a semi-circle connected by a line segment across the real axis) is easier to work with than one that encompasses only the QI pole, or one which encompasses all four poles, or is there some other reason (like that including the other poles would yield nonsense or an unhelpful identity or something)? I assume it's the former (it's just a little simpler than the alternatives) but since I can't find anything explicitly explaining it as such, I wanted to double check.


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