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Homework Help: Understand the definition of a circle in the complex plane

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data

    I know following that [tex]|z| = 1[/tex] where [tex] z \in \mathbb{C}[/tex] is the definition of unit circle in the complex plane.

    then if the exist another complex number c which lies within the distance r from z then distance from the two numbers kan be discribe as

    |z-c| = r

    If two questions here.

    Since both z and c lies within the complex plane is the distance r also a complex number?

    Can I conclude that |z-c| = r is a line within the complex plane?

    Sincerely
    Susanne

    p.s. if I square this both sides

    [tex]|z-c|^2 = r^2[/tex] then its a circle in the complex plane?
     
    Last edited: Apr 7, 2010
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  3. Apr 7, 2010 #2

    Fredrik

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    Edit: I left out a ^2 on the |z|, or a square root on the stuff after the first =, so I'm editing to fix that (after Susanne's reply below).

    [tex]|z|=\sqrt{z^*z}=\sqrt{(x-iy)(x+iy)}=\sqrt{x^2+y^2}[/itex] is a non-negative real number for every complex z (=x+iy).

    Note that [itex]|z-c|=r[/itex] is true if, and only if, [itex]|z-c|^2=r^2[/itex] is true. So it's not possible that they define different geometrical shapes.
     
    Last edited: Apr 7, 2010
  4. Apr 7, 2010 #3
    I can see that it has been a long while since I had my analysis :)

    But if I define

    [tex]z = r \cdot cos(v) + i \cdot r sin(v)[/tex] where r > 0 and [tex]v \in \mathbb{R}[/tex]

    Then modulus of z is defined |z| and if v = arg(z) then |z| = r

    and if there exist a second complex number called c

    which is defined

    [tex]c = r \cdot cos(v) + i \cdot r sin(v)[/tex]

    Then Euclidian distance between the two numbers are

    d(z,c) = |z-c| = r?

    EDIT: Do I need to show here that the two complex number do in fact lie within the same subset so I'm able to conclude that |z-c| = r? If yes do I need to go out in an epsilon-delta definition here?

    if they both belong to the same subset of [tex]\mathbb{C}[/tex]

    and the distance between the two is found by squaring on both sides of equality (according to my textbook)

    [tex]|z-c|^2 = r^2[/tex]

    and hence [tex]|z-c| = r \Leftrightarrow |z-c|^2 = r^2 [/tex]

    Sincerely
    Susanne
     
    Last edited: Apr 7, 2010
  5. Apr 7, 2010 #4

    Fredrik

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    There was mistake in my previous post. I have fixed it now, so look at it again.

    Actually, regardless of what v is, you have

    [tex]z=\sqrt{z^*z}=\sqrt{r^2\cos^2v+r^2\sin^2v}=\sqrt{r^2}=|r|[/tex]

    but if you assumed r to be positive, you have r=|r| anyway, so this makes no difference.

    That's the correct definition of distance. Just be careful not to use the same letter to mean different things. You have already used "r" in three different ways, to mean |z|, |c| and |z-c|.

    I don't understand what you're trying to do, but nothing you have mentioned so far suggests that you need to consider epsilon-delta stuff, or anything else from calculus. This is just algebra.

    I think you misunderstood something. The distance between z and c is |z-c|, as you said yourself. Also, my point when I mentioned that the two equations are equivalent, was that if we try to define a geometric shape by saying that it's the set of points for which a certain equation is true, then an equivalent equation will obviously define the same shape.
     
  6. Apr 7, 2010 #5
    I was trying to remember something from Analysis where in metric space we have

    If c is a point in [tex]\mathbb{R}^n[/tex] and that r is a given positive number. Then the set of all points x in [tex]\mathbb{R}^n[/tex] such that

    [tex]|x-c| < r[/tex] which is called an open n-ball of radius r and center c. Which is denoted B(c;r). Where B(c;r) consists of all points whose distance from c is less than r.

    and if there exists a subset T of [tex]\mathbb{R}^n[/tex] and asuming [tex]c \in T[/tex] then c is an interior point of S if there is an open ball with center at c of all points in T.

    My question is do I apply this definition to my situation and say let c be a complex number in
    [tex]\mathbb{C}^n[/tex] and let r be a positive number. Then the set of all complex numbers [tex]z \in \mathbb{C}^n[/tex] is

    [tex]|z-c| = r[/tex] and if c is a interior point of the subset T of [tex]\mathbb{C}^n[/tex] then this implies

    [tex]|z-c|^2 = r^2[/tex]??

    But if expand this I get

    [tex]|z|^2 + |c|^2 - \overline{z}c - \overline{c}z = r^2[/tex] which implies that

    [tex]|z|^2 - \overline{z}c - \overline{c}z + |c|^2 -r^2 = 0[/tex]

    and if I set [tex]d = |c|^2 -r^2[/tex] where t is a real number then

    [tex]|z|^2 - \overline{z}c - \overline{c}z + d = 0[/tex]

    Have I got this correctly?

    if yes to get the socalled generalized circle from the above the two minus signs has to change to plus and I have to multiply |z|^2 by a real number b and just to fail to realize why I must to do this?

    I multiply by a real number don't a still get a circle, but just small? depending on the size of b?
     
    Last edited: Apr 7, 2010
  7. Apr 7, 2010 #6

    Fredrik

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    It's hard to answer, because I don't know what you mean by "my situation". I don't see what exactly you're asking.

    [itex]\mathbb C^n[/itex] would be the set of ordered n-tuples of complex numbers [itex](z_1,\dots,z_n)[/itex]. We're not going to be needing any of those. Besides, you'd have to define "distance" differently in that space. The definition [tex]d((z_1,\dots,z_n),(w_1,\dots,w_n))=\sqrt{z_1^*w_1+\cdots+z_n^*w_n}[/tex] would work. (But ignore that for now, it's not relevant here).

    The set [itex]\{z\in \mathbb C|\ |z-c|=r\}[/itex] is a subset of [itex]\mathbb C[/itex], not [itex]\mathbb C^n[/itex]. Compare this to the definition of an open ball, and you'll see that this is the boundary of B(c;r), i.e. the open ball of radius r around c, in the metric space [itex]\mathbb C[/itex].

    No it would imply that there exists s>0 such that B(c;s) is a subset of T, but I don't know why you're concerned with interior points, and you really need to stop using the same letter for many different things. You also seem to be confusing [itex]\mathbb C^n[/itex] with [itex]\mathbb C[/itex].

    A complex number like c+r/2 is an interior point of B(c;r), because B(c+r/2;r/2) is a subset of B(c;r).

    I don't know why you're doing these things or why the result surprises you.
     
  8. Apr 7, 2010 #7
    I don't know either, but I am trying to understand the definition of the generalized circle in the complex plane

    mentioned here
    http://en.wikipedia.org/wiki/Generalised_circle

    but I go stuck at point mentioned in my previous post :cry:

    and to understand you correctly Fredrik then firstly if all complex numbers z in C^n then there exists a complex number c such that

    d(z,c) = |z-c|=r

    which simply implies that

    |z-c|^2 = r^2 ?

    /Susanne
     
  9. Apr 7, 2010 #8

    Fredrik

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    There are no complex numbers in [itex]\mathbb C^n[/itex]. The set of complex numbers is [itex]\mathbb C[/itex]. A sphere of radius r around a complex number c would be defined the same way as in any metric space, as [tex]\{z\in\mathbb C| d(z,c)=r\}[/tex]. The metric on [itex]\mathbb C[/itex] is defined by [itex]d(z,w)=|z-w|[/itex], so that sphere of radius r around c can be written as [tex]\{z\in\mathbb C|\ |z-c|=r\}[/tex]. Consider the special case c=0, r=1. That's the set of complex z satisfying |z|=1. If you write z=x+iy, with x and y real, that equality turns into [tex]1=|z|=\sqrt{x^2+y^2}[/tex].

    This is clearly a circle in [itex]\mathbb R^2=\mathbb C[/itex]. Note that the set [itex]\mathbb R^2[/itex] of ordered pairs of real numbers, and the set [itex]\mathbb C[/itex] of complex numbers are the same set. The field [itex]\mathbb C[/itex] is just the set [itex]\mathbb R^2[/itex] with a funny multiplication operation defined on it.

    Also note that this means that [tex]\{z\in\mathbb C|\ |z|=1\}[/tex] and [tex]\{(x,y)\in\mathbb R^2|x^2+y^2=1\}[/tex] are the same set.
     
    Last edited: Apr 8, 2010
  10. Apr 7, 2010 #9
    Okay Fredrik,

    My mistake. So my task is to loook at C^n as a field and by that obtain the correct operations to get the circle?
     
  11. Apr 8, 2010 #10

    Fredrik

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    No. Why do you think [itex]\mathbb C^n[/itex] is relevant at all?

    Do you understand why |z|=1 defines a circle?
     
    Last edited: Apr 8, 2010
  12. Apr 8, 2010 #11
    yes I understand it defines the unit circle in the complex plane! A circle or radius r = 1.

    I thought it need to show that with a circle there lies a point called c which is a complex number or do I just assume this?
     
  13. Apr 8, 2010 #12

    Fredrik

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    I'm still really confused about what you want to do and why. Do you want to prove that the set of all complex z satisfying |z|<1 isn't empty? This is the proof: |0|=0<1. No need to mention [itex]\mathbb C^n[/itex].

    You linked to a Wikipedia article earlier, but it talks about points in [itex]\mathbb C[/itex], not [itex]\mathbb C^n[/itex].
     
  14. Apr 8, 2010 #13
    I looked at my notes again and like the link yes its suppose to be [itex]\mathbb C[/itex]

    but maybe you can explain to my the it expand |z-c|^2 = r^2

    that I can change the sign in front of [itex]-c\overline{z} -\overline{c}z[/itex] to
    [itex]c\overline{z} +\overline{c}z[/itex]??? Like in the example?

    Is there some rule with addition of complex numbers and their modulus which I have forgotten? Which allows the matematician to do this?

    and why multiply |z^2| by a real number? to decrease the size of circle?

    Sincerely
    Susanne
     
  15. Apr 8, 2010 #14

    Fredrik

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    You need to be more specific. Is there a step in the derivation in this section of the generalized circle article that you don't understand?

    I don't understand why you're talking about changing the signs. And I'm not sure what example you're talking about. There's no sign change in the section I just linked to.

    I can't answer that, obviously, but there isn't much to remember. If you write z=x+iy and w=u+iv, with x,y,u,v real, can you tell me what the following are equal to?

    z*
    zw
    (zw)*
    z+z*
    z-z*
    z*z

    (Physicists usually use the asterisk (^*) instead of \overline for complex conjugation, but feel free to use \overline if you prefer).

    That step is pointless. They're trying to explain that any circle is also a generalized circle, but they're doing it badly. They should have started by defining a "generalized circle" to be a set of the form

    [tex]\{z\in\mathbb C|Az\overline z+Bz+D\overline z+D=0\}[/tex]

    where A and D are real, and C=B*. And then they should have just told you that the choices A=1, B=-c*, C=-c, D=|c|2-r2 gives you the equation of a circle of radius r around c.
     
    Last edited: Apr 8, 2010
  16. Apr 8, 2010 #15
    Hi Frederik

    What I am talking about is the step in the link where they jump from

    [tex]|z^2| - z\overline{c} - \overline{z}c + |c^2|-r^2 = 0[/tex]

    To something like

    [tex]A|z^2| + Bz + C\overline{z} + D = 0[/tex]

    is that because the author of that wiki-link is say

    that

    [tex]\overline{c} = B[/tex] discribes he complex conjugate of c

    and that [tex]c = C[/tex] and C is a complex number C

    and D = [tex]|c^2|-r^2[/tex]

    Is that what they are doing?

    Sincerely Susanne
     
  17. Apr 8, 2010 #16

    Fredrik

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    If you multiply the first equation by A, you obviously have to set [tex]B=-\overline c,\ C=-c,\ D=|c|^2-r^2[/tex] to get the second equation. The lower case c is a complex number because you started with the equation of a circle in the complex plane, with c at the center.
     
  18. Apr 9, 2010 #17
    Thank You Fredrik,

    I think I understand it now for the most part :)

    So the trick is to say

    Let [tex]c,z \mathbb{C}[/tex] and the metric distance between these two points are defined as
    [tex]d(c,z) = |z-c| = r \Leftrightarrow |z-c|^2 = r^2[/tex] which expanded gives

    [tex]|z|^2 + |c|^2 - \overline{c}z - c\overline{z} = r^2[/tex]

    Let [tex]A,D \in \mathbb{R}[/tex]

    and by setting

    [tex]B = -\overline{c}[/tex] and [tex]C = -c[/tex] [tex]D = r^2 - |c|^2[/tex] I arrive at

    which is

    [tex]A|z|^2 + Bz + C\overline{z} + D = 0, z\in \mathbb{C}[/tex] which is the generalized circle in the set of all complex numbers.

    If I set A = 1 then its a circle then its a circle in the complex plane with c as the center.

    and if A and D er zero then it a line in the complex plane and if B = 1 if C, D = 0 its just a point in the complex of radius zero.

    Is this it?

    Sincerely
    Susanne
     
  19. Apr 9, 2010 #18

    Fredrik

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    That's essentially correct, but there are some inaccuracies in what you said.

    That makes no sense. You shouldn't try to say several things at once. The distance d(c,z) between c and z is defined by d(c,z)=|c-z|. The circle of radius r around c is the set of all z that satisfies d(z,c)=r. That equation is equivalent to [itex]|z-c|^2=r^2[/itex].

    OK.

    You got the sign of D wrong. Also, the equation you end up with has A=1, so you should have mentioned that.

    If you set A=1 in an equation that defines a generalized circle, then you get a circle, but you can't tell where the center is. Your equation has C=B*=-c, and that means that the circle it represents is centered at c, but you can't "set A=1", because A is already =1.

    You should have stated the definition of a generalized circle first, and then made a comment along these lines.

    You only need to set D=0 if you want to make sure that it's a line through the origin, and...I didn't notice this until now, but the condition C=B* can't be a part of the definition of a generalized circle, because then the only straight lines that satisfy the definition of a generalized circle would be of the form Re z=constant. This means that the Wikipedia article is worse than I thought. I think any set of the form

    [tex]\{z\in\mathbb C|Az\overline z+Bz+C\overline z+D=0\}[/tex]

    where A,B,C,D are complex numbers, should be called a generalized circle.

    These choices give us [itex]A|z|^2+z=0[/itex], which is equivalent to z=-1/A*, and yes, that's a point.
     
  20. Apr 11, 2010 #19
    maybe we should write a new wiki page on subject :) One final point if I set B,C, D = 0

    then the result of the circle equation is the empty set. isn't it?

    Sincerely
    Susanne
     
  21. Apr 12, 2010 #20

    Fredrik

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    No it's not. What does the equation look like when you've set B=C=D=0? You're saying that it doesn't have any solutions, but it does.
     
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