A complex circle

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Homework Statement




[tex]\{z \in \mathbb{C}| \mathrm{\theta} \in \mathbb{R}||z| = \theta \cdot |z-1|\}[/tex]

Homework Equations



For which real-value of theta is the above no longer a circle?

The Attempt at a Solution



If I choose theta to equal zero then I get [tex]|z| = 0[/tex], but doesn't that still make it a circle? But now the zero circle??

My second attempt:

[tex]\theta \cdot |z-1| \leq r[/tex]

where r is the radius of the circle.

Then I solve this with respect to theta and get

[tex]\theta \leq r \cdot |\frac{1}{z-1}|[/tex]

My third attempt:

since theta is a real number, then [tex]\theta \in [0, \infty \}[/tex]

[tex]|z| = \sum_{\theta = 0}^\infty (\theta \cdot |z-1|) = \mathop{\lim}\limits_{\theta \to t} \int_{0}^{t} \theta |z-1| d\theta = \frac{t^2 \cdot |z-1|}{2}[/tex]

I then insert into the the inequality [tex] |z| = \frac{t^2 \cdot |z-1|}{2} \leq r[/tex]

which can be solved with respect to t:

[tex]|t| \leq \pm \sqrt{\frac{2r}{|z-1|}}[/tex]

Could somebody here please be so kind to look at my attempteds solutions for this problem, and inform me if one or any of them are correct?

Then by inserting I get that [tex] |z| = \frac{(\sqrt{\frac{2r}{|z-1|}})^2 \cdot |z-1|}{2} = r[/tex]

Best regards
Fred
 
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Answers and Replies

  • #2
HallsofIvy
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[itex]|z|= \theta |z-1|[/itex] can be interpreted as "set of all points whose distance from the orgin (|a|) is [itex]\theta[/itex] times its distance from 1 (|z-1|)". If [itex]\theta= 1[/itex] what is that set?

Another way to answer this is to "multiply it out": [itex]|z|^2= \theta^2 |z-1|^2[/itex]. If z= x+ iy then that becomes [itex]x^2+ y^2= \theta^2(x^2- 2x-1+ y^2)[/itex]. Write that quadratic in "standard form".
 
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  • #3
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[itex]|z|= \theta |z-1|[/itex] can be interpreted as "set of all points whose distance from the orgin (|a|) is [itex]\theta[/itex] times its distance from 1 (|z-1|)". If [itex]\theta= 1[/itex] what is that set?

That is the complex unit circle with radius 1.

Another way to answer this is to "multiply it out": [itex]|z|^2= \theta^2 |z-1|^2[/itex]. If z= x+ iy then that becomes [itex]x^2+ y^2= \theta^2(x^2- 2x-1+ y^2)[/itex]. Write that quadratic in "standard form".

Standard form being spherical coordinants:

[itex](r \cdot cos(\phi))^2+ (r \cdot sin(\phi))^2= \theta^2((r \cdot cos(\phi))^2- 2(r \cdot cos(\phi))-1+ (r \cdot sin(\phi))^2)[/itex]

which inturn gives

[itex]r^2 = \theta^2 \cdot (r^2 -2r \cdot cos(\phi) -1) [/itex]

Do I then solve with respect to theta in order to obtain the right real-value?

Best Regards

Fred
 
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  • #4
Dick
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Halls doesn't mean polar coordinates by 'standard form'. He means the standard form for a quadratic in two variables, ie (x-a)^2/A+(y-b)^2/B=1.
 
  • #5
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Excuse me; But English isn't my native language :)

So then I write the equation Hall wrote, and solve it with respect to theta ?

Best Regards
Fred
 
  • #6
Dick
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A quadratic written in this form (x-a)^2/A+(y-b)^2/B=1 represents a circle if A=B and A,B>0. Which value of theta will fail to give you such a form?
 
  • #7
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A quadratic written in this form (x-a)^2/A+(y-b)^2/B=1 represents a circle if A=B and A,B>0. Which value of theta will fail to give you such a form?

Hi

Then to represent a circle [tex]A=1=B=1[/tex], and [tex](1,1)>0[/tex].

Then in order for the original NOT to be a circle, then [itex]\theta \neq 1[/tex] isn't it?

Best Regards
Fred
 
  • #8
Dick
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Hi

Then to represent a circle [tex]A=1=B=1[/tex], and [tex](1,1)>0[/tex].

Then in order for the original NOT to be a circle, then [itex]\theta \neq 1[/tex] isn't it?

Best Regards
Fred

Sorry, but completely backwards. You really need to review how to recognize what kind of curve (circle, ellipse, hyperbola) a quadratic equation represents. It's not that hard, but I think you've just forgotten. In the meantime, what happens to the quadratic part of the given equation if theta=1?
 
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  • #9
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Sorry, but completely backwards. You really need to review how to recognize what kind of curve (circle, ellipse, hyperbola) a quadratic equation represents. It's not that hard, but I think you've just forgotten. In the meantime, what happens to the quadratic part of the given equation if theta=1?

Hello Dick and thank you for your reply:

I know of the equations of which you speak, and I have not forgot them. Its just then a textbook writes a equation in a certain way, it takes a bit long to understand the equation in its entirety.

The equation you present to me. Don't You mean the equation of the ellipse? Which according to an old textbook of mine is following:

((x-a)^2))/A^2 + ((x-b)^2)/B^2 = r, where (A,B)>0

Then if A and B both equal 1, then its a circle, but if (A,B) > 0, then it an ellipse. Then this must apply to my original problem too. That if theta equals 1, then its a circle, but if theta > 0 then its a ellipse of an unspecified size. That is what you refering too? Isn't??

Best Regards
Fred
 
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  • #10
Dick
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If A and B both equal 1, it's a circle. If A and B are equal and nonzero, it's STILL a circle even if they are not equal 1. I don't know what an 'ellipse of infinite size' is - do you mean parabola? Maybe it's just language but I don't think you remember this stuff as well as you think you do. Once more, what is the quadratic like if theta equals one? Don't guess. Write it down.
 
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  • #11
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If A and B both equal 1, it's a circle. If A and B are equal and nonzero, it's STILL a circle even if they are not equal 1. I don't know what an 'ellipse of infinite size' is - do you mean parabola? Maybe it's just language but I don't think you remember this stull as well as you think you do. Once more, what is the quadratic like if theta equals one? Don't guess. Write it down.


Once again taking

[itex]|z|^2 = {\theta}^2 \cdot |(x+iy)-1|^2[/itex]

this is the same as, (but I get a different right side, than Hall)

[itex]x^2 + y^2 = {\theta}^2(x^2 + y^2 - 2x + 1)[/itex]

which in sf gives:

[itex]x^2+y^2 = \theta^2 ((x-1)^2 + y^2 + 1)[/itex]

But then if theta = 1, then I get that the circle with a the zenter in (0,0) equals a circle with center in (1,0), but who has a negative radius. That cannot be right can it?
 
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  • #12
Dick
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Once again taking

[itex]|z|^2 = {\theta}^2 \cdot |z-1|^2[/itex]

this is the same as, (but I get a different right side, than Hall)

[itex]x^2 + y^2 = {\theta}^2(x^2 + y^2 - 2x + 1)[/itex]

which in sf gives:

[itex]x^2+y^2 = \theta^2 ((x-1)^2 + y^2 + 1)[/itex]

Ok, you've corrected Halls. You are being careful. Now do two things. Put theta equal to 1 and tell me what kind of curve you get - then think about what happens if theta is not one.
 
  • #13
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Ok, you've corrected Halls. You are being careful. Now do two things. Put theta equal to 1 and tell me what kind of curve you get - then think about what happens if theta is not one.

If re-arrange it, then I get a straight line: 2x-2. How does that sound?

Best Regards
Fred
 
  • #14
Dick
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No, you get 2x-1=0. I thought you were being careful!? And makes sense. It's the set of points equidistant from (0,0) and (1,0). Now can you convince yourself for other values of theta you get a circle?
 
  • #15
HallsofIvy
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[itex]|z|= \theta |z-1|[/itex] can be interpreted as "set of all points whose distance from the orgin (|a|) is [itex]\theta[/itex] times its distance from 1 (|z-1|)". If [itex]\theta= 1[/itex] what is that set?

That is the complex unit circle with radius 1.
The set of points equidistant from two fixed points is a straight line! It is the perpendicular bisector of the line segment between the two points.



Standard form being spherical coordinants:
Why in the world would "standard form" for a 2 dimensional figure involve spherical coordinates?
 
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  • #16
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I am sorry but are Y sure about the 2x-1 = 0.

Because I get ???

[tex]x^2 + y^2 - (x^2 -2x + 1) - y^2 -1 = 2x - 2 = 0.[/tex]

So basicly I conclude here that if [tex]\theta = 1[/tex] then [tex]|z| = |z-1| [/tex]is not a circle. But If [tex]\theta > 1[/tex] then it return to be a circle!

I have a second follow up question which also deals with analysis.

Given [itex]\{z \in \mathbb{C}| e^{|iz|} \leq 1 \}[/itex],

Determ the set for the above:

By the Euler thereom then:

[tex]e^{|iz|} = |cos(z) + i sin(z)| \leq 1[/tex]

then if z = x+iy I get

[tex]e^{|iz|} = |cos(x+iy) + isin(x+iy)| \leq 1 [/tex]

which inturn gives me

[tex]e^{-y} \leq 1[/tex]

But doesn't say nothing much about the set. Does it?

Best Regards
Fred
 
  • #17
Dick
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Where did you get the second '1' in your equation starting from this [itex]x^2+ y^2= \theta^2(x^2- 2x+1+ y^2)[/itex]??
 
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  • #18
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No where you answer is correct. The extra "1" was dumb a** mistake on my part. Sorry :)

Best Regards

Fred

p.s. Regarding my second question. That I am able to reduce it e^-y <= 1.

Then "y" is the img-part of the number z = x+yi. Then the set to describe the equation e^{|iz|} <= 1, is that the set of all complex number on the form z = i?
 
  • #19
Dick
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On your second problem - there are so many things wrong here that it's hard to know where to start. Start from scratch and if you write something down give some justification for why you think it's true. Hint: |iz| is a real number. What kind?
 
  • #20
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Okay here is the question again in full:

Deduce the set

[tex]\{z \in \mathbb{C}|e^{|iz|} \leq 1 \}[/tex]

Solution:

Originally I thought that I needed to write out the above using Eulers formula, but since it doesn't make any difference. I choose not too.

Dick, You say that z should be scene as a real number. Then the z you are refering to must z = x+i0, there by where x belong to R. Isn't it?

Best Regards
Fred
 
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  • #21
Dick
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Ok, start to deduce the set. What can you tell me about |iz|?
 
  • #22
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Ok, start to deduce the set. What can you tell me about |iz|?


Now there is something funny happing if I tap in |iz| where z = x+iy. I get

[tex]|iz| = \sqrt{(x^2+y^2)}[/tex].

Doesn't this then mean, that express a circle with radius r and which has the center in (0,0)??

Best Regards
Fred
 
  • #23
Dick
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Now there is something funny happing if I tap in |iz| where z = x+iy. I get

[tex]|iz| = \sqrt{(x^2+y^2)}[/tex].

Doesn't this then mean, that express a circle with radius r and which has the center in (0,0)??

Best Regards
Fred

Yes, it would. So what can you tell me about r.
 
  • #24
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Yes, it would. So what can you tell me about r.


I can say that r is the real-valued expression of z.

Then to deduce the set, is it then

[tex]\{r = \sqrt{x^2+y^2}|x,y,r \in \mathbb{R}\}[/tex] ??

Best Regards

Fred
 
  • #25
Dick
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No. You've decided that exp(|iz|)=exp(r) where r=sqrt(x^2+y^2). How can that be less than or equal to 1?
 

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