Understand the definition of a circle in the complex plane

  • Thread starter Susanne217
  • Start date
  • #1
Susanne217
317
0

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?
 
Last edited:

Answers and Replies

  • #2
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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:
  • #3
Susanne217
317
0
[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
 
Last edited:
  • #4
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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
Susanne217
317
0
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?
 
Last edited:
  • #6
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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
Susanne217
317
0
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
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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:
  • #9
Susanne217
317
0
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
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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:
  • #11
Susanne217
317
0
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
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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
Susanne217
317
0
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
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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.
 
Last edited:
  • #15
Susanne217
317
0
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
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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
Susanne217
317
0
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
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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
Susanne217
317
0
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
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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
Susanne217
317
0
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 :(
 
Last edited:
  • #22
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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
Susanne217
317
0
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
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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?
 
Last edited:
  • #25
Susanne217
317
0
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?
 
  • #26
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420
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?
I didn't mean that you should tell me what shape a generalized circle is for different choices of A,B,C,D. I want you to think about what the notation means. Forget about the equation for the moment. Do you understand for example what [tex]\{x\in \mathbb R|x^2=1\}[/tex] means? (And please don't say that it means that x is 1 or -1. I'm not asking you to solve the equation. I'm asking you tell me what the notation means).

The choice A=B=C=D=0 does not give you the empty set.

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?
I don't even know what the word "that" refers to in the first sentence. Sounds like you're referring to the text I quoted, but what you're saying doesn't answer the questions I asked. Here they are again: Why can you choose A to be real? Why does A being real imply that D is real too when the generalized circle is a circle?
 
  • #27
Susanne217
317
0
I didn't mean that you should tell me what shape a generalized circle is for different choices of A,B,C,D. I want you to think about what the notation means. Forget about the equation for the moment. Do you understand for example what [tex]\{x\in \mathbb R|x^2=1\}[/tex] means? (And please don't say that it means that x is 1 or -1. I'm not asking you to solve the equation. I'm asking you tell me what the notation means).

The choice A=B=C=D=0 does not give you the empty set.


I don't even know what the word "that" refers to in the first sentence. Sounds like you're referring to the text I quoted, but what you're saying doesn't answer the questions I asked. Here they are again: Why can you choose A to be real? Why does A being real imply that D is real too when the generalized circle is a circle?

The first question is the set of real numbers which satisfies the equation x^2 = 1.

Sorry, about that my english gramma isn't very good....

What I don't understand about you question(sorry for me being dumb) if choose "a" to be real then he whole a|z|^2 and becomes a point in real plane by itself from what I see it..

However if choose a to be complex then a|z|^2 lies within the complex plane and hence by itself lies within the complex plan.
Thats the only difference that I can see at the moment :confused:

regarding the second question if B=C=D=0 then the point generated by the eqn is zero and the only element which is contained within the empty set is zero and thus its also the empty set. Can I put it like that?
 
  • #28
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420
The first question is the set of real numbers which satisfies the equation x^2 = 1.
Yes, that's exactly what I had in mind. In general, the notation [tex]\{x\in S|P\}[/tex] means "the set of all x in the set S, for which the statement P is true". Now consider the set [tex]\{z\in\mathbb C|Az\overline z+Bz+C\overline z+D=0\}[/tex] with A=B=C=D=0 again. (Edit: For some reason I wrote A=B=C=C instead of A=B=C=D=0 above. I have fixed it now, after you replied). What is the statement "P" in this case? For which values of z is that statement true?

What I don't understand about you question(sorry for me being dumb) if choose "a" to be real then he whole a|z|^2 and becomes a point in real plane by itself from what I see it..
Now you seem to be talking about choosing A real after choosing B=C=D=0. That's not what we were discussing. You were supposed to consider an arbitrary generalized circle with A,B,C,D complex, and realize that there's always a generalized circle with A real that is actually the same set. Can you see why?

However if choose a to be complex then a|z|^2 lies within the complex plane and hence by itself lies within the complex plan.
The solution to A|z|2=0 is the same regardless of A. It doesn't matter if A is real, as long as it's not =0.

regarding the second question if B=C=D=0 then the point generated by the eqn is zero and the only element which is contained within the empty set is zero and thus its also the empty set. Can I put it like that?
With the assumptions you have made (you weren't supposed to set B=C=D=0, but I'll play along for the moment), it's correct that the only solution to the equation (which has been reduced to A|z|2=0 by your assumption) is z=0.

Maybe it's the language barrier, but it looks like you just said that the empty set contains 0. :confused: A set that contains something isn't empty. {0} is not equal to {}.
 
Last edited:
  • #29
Susanne217
317
0
Yes, that's right. In general, the notation [tex]\{x\in S|P\}[/tex] means "the set of all x in the set S, for which the statement P is true". Now consider the set [tex]\{z\in\mathbb C|Az\overline z+Bz+C\overline z+D=0\}[/tex] with A=B=C=C again. What is the statement "P" in this case? For which values of z is that statement true?

Thats the set of all complex numbers which satisfies the eqn for the generalized circle.

Now you seem to be talking about choosing A real after choosing B=C=D=0. That's not what we were discussing. You were supposed to consider an arbitrary generalized circle with A,B,C,D complex, and realize that there's always a generalized circle with A real that is actually the same set. Can you see why?

yes cause if I a to be real its still on the form a = x + i0 and that that it lies within the same plane if a = x+iy.

With the assumptions you have made (you weren't supposed to set B=C=D=0, but I'll play along for the moment), it's correct that the only solution to the equation (which has been reduced to Az=0 by your assumption) is z=0.

Maybe it's the language barrier, but it looks like you just said that the empty set contains 0. :confused: A set that contains something isn't empty. {0} is not equal to {}.

Maybe it is but in order for the eqn to represent the empty then I have to show that there exists a form of the eqn where there is no solution? Hence the empty set?

It can't be simply that any z which lies outside the circle represents the empty set?
 
  • #30
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420
Thats the set of all complex numbers which satisfies the eqn for the generalized circle.
And what is that equation in this case? (Sorry if my typo confused you. It's supposed to be A=B=C=D=0, not A=B=C=C. I have edited my post to correct the mistake). For which values of z is the equation true?

yes cause if I a to be real its still on the form a = x + i0 and that that it lies within the same plane if a = x+iy.
You're just saying that [itex]\mathbb R\subset\mathbb C[/itex]. That doesn't answer the question at all.

Maybe it is but in order for the eqn to represent the empty then I have to show that there exists a form of the eqn where there is no solution? Hence the empty set?
I have told you lots of times that you won't end up with the empty set.

It can't be simply that any z which lies outside the circle represents the empty set?
I don't know why you think points in the complex plane have anything to do with the empty set.
 
  • #31
Susanne217
317
0
And what is that equation in this case? (Sorry if my typo confused you. It's supposed to be A=B=C=D=0, not A=B=C=C. I have edited my post to correct the mistake). For which values of z is the equation true?

Iff A=B=C=D=0 then the z which statisfies this instance of the original equation is z = 0 as far as I see it such that

0*|0|^2 + 0*0 + 0*0 + 0 = 0

But then again if I choose any complex number z and hold A=B=C=D=0 then it will always satify the original equation. Damn it! It must :frown:

You're just saying that [itex]\mathbb R\subset\mathbb C[/itex]. That doesn't answer the question at all.

It must be something along the lines. If I choose A to be an abitrary complex number or choose A to be abitrary real number then it both cases it still has to satify the conditions of the original equation

A|z|^2 + B*^z + Cz + D = 0. (1)

And it simply has to be it doesn't matter if you choose A to be complex or real....

I have told you lots of times that you won't end up with the empty set
I don't know why you think points in the complex plane have anything to do with the empty set.

I found some old notes which claims that (1) can be seen as both a straight line, circle and as the empty set.
We have been able to show the line and the circle, but I am simply unable to wrap my head around that its possible to see (1) as the empty set. Because as I see the empty set its like circle which contains no elements and has no radius.

Can it be a mistake in the notes? The part with the empty set?
 
  • #32
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420
Iff A=B=C=D=0 then the z which statisfies this instance of the original equation is z = 0 as far as I see it such that

0*|0|^2 + 0*0 + 0*0 + 0 = 0

But then again if I choose any complex number z and hold A=B=C=D=0 then it will always satify the original equation. Damn it! It must :frown:
I don't understand the first comment (about z=0) at all, because you have correctly determined the equation to be 0=0. I don't know how you got z=0 from that. The set we're talking about is [tex]\{z\in\mathbb C|0=0\}[/tex], and if you understand what the notation means, you must know what set this is.

It must be something along the lines. If I choose A to be an abitrary complex number or choose A to be abitrary real number then it both cases it still has to satify the conditions of the original equation

A|z|^2 + B*^z + Cz + D = 0. (1)

And it simply has to be it doesn't matter if you choose A to be complex or real....
It sounds like you have already lost track of what you were supposed to answer. I'll ask the question in another way, which probably makes this too easy, but it seems that we're not making much progress.

Let A,B,C,D be complex numbers, and define [tex]S=\{z\in\mathbb C|Az\overline z+Bz+C\overline z+D=0\}[/tex]. Your task is to find a way to express the same set S in the form [tex]\{z\in\mathbb C|A'z\overline z+B'z+C'\overline z+D'=0\}[/tex], where A',B',C',D' are complex numbers, and Im A'=0.

I found some old notes which claims that (1) can be seen as both a straight line, circle and as the empty set.
We have been able to show the line and the circle, but I am simply unable to wrap my head around that its possible to see (1) as the empty set. Because as I see the empty set its like circle which contains no elements and has no radius.

Can it be a mistake in the notes? The part with the empty set?
I haven't seen those notes, so it's hard to tell. How exactly do the notes define a generalized circle?
 
  • #33
Susanne217
317
0
I don't understand the first comment (about z=0) at all, because you have correctly determined the equation to be 0=0. I don't know how you got z=0 from that. The set we're talking about is [tex]\{z\in\mathbb C|0=0\}[/tex], and if you understand what the notation means, you must know what set this is.


A set which contains [tex]\{z\in\mathbb C|0=0\}[/tex] that must be a set which contains no elements and thats the empty set. :surprised

It sounds like you have already lost track of what you were supposed to answer. I'll ask the question in another way, which probably makes this too easy, but it seems that we're not making much progress.

I properly have :)

Let A,B,C,D be complex numbers, and define [tex]S=\{z\in\mathbb C|Az\overline z+Bz+C\overline z+D=0\}[/tex]. Your task is to find a way to express the same set S in the form [tex]\{z\in\mathbb C|A'z\overline z+B'z+C'\overline z+D'=0\}[/tex], where A',B',C',D' are complex numbers, and Im A'=0.

If B=C=D = 0 and A is real that I get that A'z =0

I haven't seen those notes, so it's hard to tell. How exactly do the notes define a generalized circle?

They simply say that if you let A,D be real numbers and B, C be complex then one gets equation above which be viewed as both a circle next a straight and then the empty set...
 
  • #34
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420
A set which contains [tex]\{z\in\mathbb C|0=0\}[/tex] that must be a set which contains no elements and thats the empty set. :surprised
I have to be honest, I don't think I've ever seen a statement that's as incorrect as this one. I mean, to say that 3 is an even number would be incorrect, but this is much, much worse than that. Not only is the conclusion 100% false, but the argument that you used to obtain the conclusion is logically inconsistent. I don't know why you think a set can be both empty and non-empty. Please try again. You need to think about what the notation means, and actually use that knowledge.

Also, we were talking about the set [tex]\{z\in\mathbb C|0=0\}[/tex]. I have no idea why you suddenly started talking about a set that contains [tex]\{z\in\mathbb C|0=0\}[/tex] (and at the same time doesn't).

If B=C=D = 0
When you've been given four complex numbers, you can't choose three of them to be zero. Please try again. I was very careful to state the question in a way that I thought couldn't be misunderstood. Please read the question carefully (in my previous post).
 
  • #35
Susanne217
317
0
I guess I hit my head with my Old Calculus book and thats to blame for this statement ;)

Off cause since it contains the zero element then the set isn't empty, sorry about that Fredrik.

I hope you can forgive me :)

But how in anyones name a the function can be claimed to be the empty set that I have totally no idear on how it can be shown.

Can it be a trick someone wrote into the conclusion regarding the generalized circle.

Only way that the eqn of generalized circle to represent the empty set is for the domain of eqn to be empty, and as understand your thoughts Fredrik then the domain of generalized circle can never be empty. Thus it can't be show to represent the empty set?
 

Suggested for: Understand the definition of a circle in the complex plane

Replies
7
Views
461
Replies
4
Views
142
Replies
5
Views
268
  • Last Post
Replies
6
Views
198
Replies
6
Views
206
Replies
2
Views
178
Replies
14
Views
466
Replies
2
Views
326
Top