Understand the definition of a circle in the complex plane

[Fredrik]

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

I hope you can forgive me :)

Of course. But you're still nowhere near the very obvious and very trivial truth about what set $$\{z\in\mathbb C|0=0\}$$ is.

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.

I think you should focus on the two problems I've been trying to get you to solve first. Both of them are about as easy as a math problem can get. As long as you haven't understood them, I don't think you can understand any problems of this sort, including the question of whether it's possible for a generalized circle to be the empty set.

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?

A "domain" is a set on which a function is defined. You probably mean the set of solutions of the equation. If that's empty, then the generalized circle is the empty set. Is that possible? I actually haven't thought that through, so at this precise moment, I can't tell you the answer. We can return to that when you have solved the two trivial problems.

Edit: I have to add that I think you should try to re-evaluate your whole approach to solving math problems. It seems to me that most of the time, you try to guess the solution without actually using the information in the problem you have been given. Math doesn't work that way. I don't think anything does, but there's no other field where it's as important to use the information you've been given as in mathematics.

 

[Susanne217]

Of course. But you're still nowhere near the very obvious and very trivial truth about what set $$\{z\in\mathbb C|0=0\}$$ is.


I think you should focus on the two problems I've been trying to get you to solve first. Both of them are about as easy as a math problem can get. As long as you haven't understood them, I don't think you can understand any problems of this sort, including the question of whether it's possible for a generalized circle to be the empty set.


A "domain" is a set on which a function is defined. You probably mean the set of solutions of the equation. If that's empty, then the generalized circle is the empty set. Is that possible? I actually haven't thought that through, so at this precise moment, I can't tell you the answer. We can return to that when you have solved the two trivial problems.

Edit: I have to add that I think you should try to re-evaluate your whole approach to solving math problems. It seems to me that most of the time, you try to guess the solution without actually using the information in the problem you have been given. Math doesn't work that way. I don't think anything does, but there's no other field where it's as important to use the information you've been given as in mathematics.

The problem here is that the problem is generalized I personally find it difficult to approach a problem without any specific numbers. That maybe makes me stupid, I guess.

As I said I understand the circle as it is and the straight line, but not the other things that's hard. Is it possible to say if one chooses z's which doesn't satisfy the equation, then in that case the solution set for the equation is empty?
If not why not?

 

[Fredrik]

The problem here is that the problem is generalized I personally find it difficult to approach a problem without any specific numbers. That maybe makes me stupid, I guess.

No, I don't think so, but it definitely means that you still haven't found the right way to approach mathematical problems, and that you need to keep looking for it. You also need to recognize when you have been given specific numbers that just happen to be represented by letters. For example, consider one of the problems I gave you:

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

This is a problem where you've been given four complex numbers A,B,C,D. Does it really matter what values they represent? All complex numbers obey the same rules (like AB=BA) anyway. (The only exception is that you can't divide by 0). When you tried to solve this extremely easy problem, you kept changing the information you had been given (by specifying values of A,B,C,D that you hadn't been given by me).

As I said I understand the circle as it is and the straight line, but not the other things that's hard. Is it possible to say if one chooses z's which doesn't satisfy the equation, then in that case the solution set for the equation is empty?
If not why not?

Can you really say something about the set of solutions to an equation by considering a value of the variable that doesn't satisfy the equation? For example, does the choice x=117 say anything about the solutions of 3x=6?

This is exactly the sort of thing I'm talking about. If you want to find the solutions to an equation, you need to use the equation you've been given, but every time you approach a problem, you ignore the information you've been given (a mistakes that guarantees failure) and usually change the information to something that you think suits you better. I will tell you the solution to the other problem I gave you. The question was this: What set is $$\{z\in\mathbb C|Az\overline z+Bz+C\overline z+D=0\}$$ when A=B=C=D=0. What follows is the solution:

The notation means "the set of all complex numbers z such that the equation is satisfied". When A=B=C=D=0, that equation is 0=0, so we're looking for $$\{z\in\mathbb C|0=0\}$$, i.e. the set of all complex numbers such that the equality 0=0 is true. Since there are no values of z that would make 0≠0, the equality is true for all z. So the answer is

$$\{z\in\mathbb C|0=0\}=\mathbb C$$

Now, was that so hard?

The fact that you couldn't come up with this on your own shows very clearly that you're not thinking about what the symbols (and the words) mean. You kept saying that this set must be empty, sometimes even after saying that it isn't empty. Consider this. To say that a set S is empty means that the statement "S is empty" is true. To say that S is not empty means that the same statement is false. So at least twice you attributed the values "true" and "false" to the same statement. Nothing is ever both true and false in mathematics. You need to remember that, and apply it to every problem you try to solve.

You were clearly not thinking about what your own answer meant either. Every time you said that "the set of all z such that 0=0" is something other than [tex]\mathbb C[/itex], you were actually saying that there exists a complex number z such that 0≠0. Either you didn't think at all about what you were saying, or you haven't understood what an equality sign means. If x and y are numbers, x=y means that they're the same number. If X and Y are sets, X=Y means that they have the same members.

Think about this for a second: Why are we allowed to add -2 to both sides of the equation 3x+2=14? (Think about it before you move your mouse over the hidden text below).

Spoiler

Because the equality sign means that you have the same number on both sides. In this case the equality represents the statement "14 is the same number as 14". You can add -2 because it gives you a new equality, which represents the statement "14-2 is the same number as 14-2".

You also don't seem to understand what an equation is. When you're asked to "solve the equation" 3x+2=14 in a math book about real numbers, you're actually being asked to "find all the real numbers that you can substitute for x without making the equality false". I have to assume that you don't understand this, since you're asking if choosing a z that doesn't satisfy $$Az\overline z+Bz+C\overline z+D=0$$ can tell you which values of z that do satisfy the equation.

The fact that you haven't been able to solve the problem I quoted earlier in this post strongly suggests that you don't understand what the notation means, or at least that you haven't thought about it. Please try to solve it again. It's extremely easy for a person who understands the notation, who understands what an equation is, and who knows anything about complex numbers.

 

[Susanne217]

Maybe you think this is dumb Fredrik, but I will try approach the problem by choosing some numbers

$$A = 1+i$$ and $$B = -(2-i)$$ and $$C = -(2+i)$$ and $$D = (2+i)^2 - 4$$

Iff I inserted these values into the equation for the circle I get

$$(1+i)\cdot |z|^2 + (-2-i))z + (-2+i) \cdot \overline{z} +((2+i)(2-i) - 4) = 0$$

where $$z \in \mathbb{C}$$

Which is a circle in the complex plane of radius 2 and which centers in c = (2+i)

Taking away D by setting D = 0

Then the solution is a line in the complex plane.

and then by presenting that

$$(1+i)|z|^2 = 0$$

then its now a point name z = 0.

and if A to be a real number $$2|z|^2 = 0$$ then it will still only have the solution z = 0 which is a point.

and if I choose A, B, C be zero and D = 0 - 2^2 -> D = -4

then I end up with

$$0\cdot |z|^2 + 0\cdot z + 0 \cdot \overline{z} + (0-2^2) = 0$$

that statement will always be false and hence the empty set.

Spoiler

p.s. What is basicly curriculum at one University in one country can as you know differ from other countries Fredrik. So what you were thought to be basic stuff then you went to University can easily be something me as a Undergraduate Student have not seen yet. So no need to think that I can't solve the Pre-Highschool level problems. And just as you know it I do know that that $$x \in \mathbb{R}$$ means that x belongs to the set of realnumbers and $$z \in \mathbb{C}$$ that z belongs to the set of complex numbers. That said I am learning by the stuff you have been writting become a better student of mathematics. And learning any new ways to approach things I'm very gratefull for.

 

[Fredrik]

Maybe you think this is dumb Fredrik, but I will try approach the problem by choosing some numbers

$$A = 1+i$$ and $$B = -(2-i)$$ and $$C = -(2+i)$$ and $$D = (2+i)^2 - 4$$

It's certainly OK to try some numbers just to see what you get.

Iff I inserted these values into the equation for the circle

That's not what you're doing. You're inserting them into the equation that defines a generalized circle.

I get

$$(1+i)\cdot |z|^2 + (-2-i))z + (-2+i) \cdot \overline{z} +((2+i)(2-i) - 4) = 0$$

where $$z \in \mathbb{C}$$

Which is a circle in the complex plane of radius 2 and which centers in c = (2+i)

No, it's not. This equation is not equivalent to the equation that defines a circle of radius 2 around c. Try writing down that equation and comparing it to what you've got.

and if I choose A, B, C be zero and D = 0 - 2^2 -> D = -4
...
that statement will always be false and hence the empty set.

Yes, this is correct. $$\{z\in\mathbb C|Az\overline z+Bz+C\overline z+D=0\}$$ is empty when A=B=C=0 and D≠0.

So no need to think that I can't solve the Pre-Highschool level problems. And just as you know it I do know that that $$x \in \mathbb{R}$$ means that x belongs to the set of realnumbers...

I don't doubt that you can solve pre-high school problems, but the mistakes you're making are very elementary. Some of them almost at a pre-preschool level. So I don't know where to begin.

If you do understand the notation and extremely basic stuff like what an equality means, then why weren't you able to see that $$\{z\in\mathbb C|0=0\}=\mathbb C$$? And why haven't you been able to solve the other problem, which is just as easy? I'll tell you the solution: Multiply both sides of the equation with A*. Can you at least tell me why that solves the problem?