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Understand the definition of a circle in the complex plane

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Homework Statement



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?
 
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Answers and Replies

  • #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.
 
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  • #3
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[itex]|z|=z^*z=(x-iy)(x+iy)=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.
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
 
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  • #4
Fredrik
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There was mistake in my previous post. I have fixed it now, so look at it again.

[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
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.

Then Euclidian distance between the two numbers are

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

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]
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.

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]
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.
 
  • #5
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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.
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?
 
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  • #6
Fredrik
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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.
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).

Then the set of all complex numbers [tex]z \in \mathbb{C}^n[/tex] is

[tex]|z-c| = r[/tex]
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].

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]??
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).

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]t = |c|^2 -r^2[/tex] where t is a real number then

[tex]|z|^2 - \overline{z}c - \overline{c}z + t = 0[/tex]
I don't know why you're doing these things or why the result surprises you.
 
  • #7
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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.
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
 
  • #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.
 
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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?
 
  • #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?
 
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  • #11
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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?
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?
 
  • #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].
 
  • #13
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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].
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
 
  • #14
Fredrik
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but maybe you can explain to my the it expand |z-c|^2 = r^2
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?

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?
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.

Is there some rule with addition of complex numbers and their modulus which I have forgotten?
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).

and why multiply |z^2| by a real number? to decrease the size of circle?
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.
 
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  • #15
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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
 
  • #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.
 
  • #17
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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.
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
 
  • #18
Fredrik
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That's essentially correct, but there are some inaccuracies in what you said.

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]
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].

which expanded gives

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

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

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

and if A and D er zero then it a line in the complex plane
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.

and if B = 1 if C, D = 0 its just a point in the complex of radius zero.
These choices give us [itex]A|z|^2+z=0[/itex], which is equivalent to z=-1/A*, and yes, that's a point.
 
  • #19
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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.
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
 
  • #20
Fredrik
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One final point if I set B,C, D = 0

then the result of the circle equation is the empty set. isn't it?
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.
 
  • #21
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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.
I can see that now then its still a point in the complex plane. Then A=B=C=D=0 is the only way for it to become the empty set?

Because if I set

a = 0 (which is a real number)

B = 0+i0 (Which is a complex number)

C = 0+i0

and r = 0

which implies that D = 0

then the only solution must be the empty set :(
 
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  • #22
Fredrik
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You're right that B=C=D=0 defines a point, the point 0 to be exact.

It seems that you're still not really thinking about what the definition of the generalized circle means. We defined a generalized circle to be a 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. Now consider specifically the "generalized circle" with A=B=C=D=0. What set is that? (You really need to think about what the definition above means).

This actually means that I didn't get the definition quite right. We don't want this set to be called a generalized circle, so the definition should say that a generalized circle is a set of that form, with A,B,C,D complex numbers that are not all zero.
 
  • #23
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You're right that B=C=D=0 defines a point, the point 0 to be exact.
Good that I understand :)

It seems that you're still not really thinking about what the definition of the generalized circle means. We defined a generalized circle to be a 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. Now consider specifically the "generalized circle" with A=B=C=D=0. What set is that? (You really need to think about what the definition above means).

This actually means that I didn't get the definition quite right. We don't want this set to be called a generalized circle, so the definition should say that a generalized circle is a set of that form, with A,B,C,D complex numbers that are not all zero.
As I understand if I define A=B=C=D=0 then the set contains no elements and hence its with those conditions the empty set.

Finally, I thought that A and D were suppose to be considered as real numbers? Only B and C complex?

Susanne
 
  • #24
Fredrik
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As I understand if I define A=B=C=D=0 then the set contains no elements and hence its with those conditions the empty set.
This is wrong. What does the choice A=B=C=D=0 do to the mathematical expression [tex]\{z\in\mathbb C|Az\overline z+Bz+C\overline z+D=0\}[/tex]? What does that expression look like after you've set all those numbers to zero? (You really need to think about what the notation means. It helps to express it in words instead of symbols).

Finally, I thought that A and D were suppose to be considered as real numbers? Only B and C complex?
No, they can all be complex. But note that A can be chosen to be real without loss of generality. Do you see why? When A is real and the generalized circle is a circle, then D is real too. Do you see why?
 
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  • #25
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This is wrong. What does the choice A=B=C=D=0 do to the mathematical expression [tex]\{z\in\mathbb C|Az\overline z+Bz+C\overline z+D=0\}[/tex]? What does that expression look like after you've set all those numbers to zero? (You really need to think about what the notation means. It helps to express it in words instead of symbols).
looking at the notation. B = -*^c and C = -c and D = |c|^2-r^2

that means that if A = B = C = D = 0 then the equation doesn't represent a circle, line or point and if the multiplication of A with (z times *^z) that gives us a complex us real rumber times a complex number which is still a complex number, and if A is zero then that still gives us a complex number om the form 0+i0 and that just represents a point in both the complex set and real set. If D is zero that that means that both c and its positive and negative conjugate become zero and they still represent a point zero in both the complex and real plane. Then in order to making the equation represent the empty set do I have prove that it contains no elements and thusly is the empty set?

No, they can all be complex. But note that A can be chosen to be real without loss of generality. Do you see why? When A is real and the generalized circle is a circle, then D is real too. Do you see why?
All I can see here is that result in the same representation of the equation as a line, circle and point. Is that what you are refering to Fredrik?
 

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