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Is this true?? (1 Viewer)
- Thread starter no_arg
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darkliight
I ponder, weak and weary
I'm not sure what you're asking ... say your fixed circle has radius r, do you mean ||z|| = r or ||z|| <= r ?
If the former, it's true, if the later, it's false.
If the former, it's true, if the later, it's false.
More precisely
suppose that |z-a|=r (z lies on a circle of radius r centred at a) and |a| is not equal to r (circle does not contain 0)
prove that
the locus of 1/z is also a circle.
Its trivial of course when a=0
I'm after a variety of different proofs (no insane determinant arguments please!)
suppose that |z-a|=r (z lies on a circle of radius r centred at a) and |a| is not equal to r (circle does not contain 0)
prove that
the locus of 1/z is also a circle.
Its trivial of course when a=0
I'm after a variety of different proofs (no insane determinant arguments please!)
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If I recall correctly, the question is equally as interesting (and perhaps just as difficult) if you instead create the constraint that the circle must pass through the origin.
From memory, in this case, the locus of 1/z will actually be a straight line.
Disclaimer: I may not recall correctly.
From memory, in this case, the locus of 1/z will actually be a straight line.
Disclaimer: I may not recall correctly.
Correct. Someone asked this before. I put 2 solutions and several others put up solutions also. Here it is:
http://community.boredofstudies.org/14/mathematics-extension-2/104616/complex-numbers-question.html
http://community.boredofstudies.org/14/mathematics-extension-2/104616/complex-numbers-question.html
This is not a proof, but just a way to confirm it. Let z=x+iy, then 1/z = x-iy, which is the conjugate of z, ie the reflection of z in the real axis. Therefore all the points representing z that lie on that fixed circle will be reflected in the real axis to form another circle.
Raginsheep
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Isn't that only when mod z is 1?
Well if the original circle is |z-a|=r, I got (again using determinants) that the new one has centre conj(a)/(|a|<sup>2</sup>-r<sup>2</sup>) and radius r/||a|<sup>2</sup>-r<sup>2</sup>| - but it is not necessary to find these things in order to answer your original question.no_arg said:
It can be generalised to linear fractional transformation f(z)=(az+b)/(cz+d) whereby circles not through the origin are mapped to circles, and if they pass through the origin, they are mapped to lines:
http://mathworld.wolfram.com/LinearFractionalTransformation.html
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Lets look at an easier problem.
Suppose 2x<sup>3</sup>-5x<sup>2</sup>+4x+6=0 has roots α, β, γ. Find an equation with roots 1/α, 1/β, 1/γ.
Solution.
Replace x with 1/x.
∴ 2/x<sup>3</sup>-5/x<sup>2</sup>+4/x+6=0.
∴ 6x<sup>3</sup>+4x<sup>2</sup>-5x+2=0.
x is a dummy variable. That's why we can say "replace x with 1/x". It's the same with locus.
But if you don't like saying "replace x with 1/x", you can instead do it like this:
Let y=1/x.
∴ 2/y<sup>3</sup>-5/y<sup>2</sup>+4/y+6=0.
∴ 6y<sup>3</sup>+4y<sup>2</sup>-5y+2=0.
The result is the same, with dummy variable y instead of x.
It's the same in locus problems. In my syllabus solution, instead of saying "replace z with 1/z", you can say "Let w=1/z" and get the same result, i.e, the same circle with the same centre and same radius, but with a dummy variable w instead of z in the equation.
Hope that helps.
Suppose 2x<sup>3</sup>-5x<sup>2</sup>+4x+6=0 has roots α, β, γ. Find an equation with roots 1/α, 1/β, 1/γ.
Solution.
Replace x with 1/x.
∴ 2/x<sup>3</sup>-5/x<sup>2</sup>+4/x+6=0.
∴ 6x<sup>3</sup>+4x<sup>2</sup>-5x+2=0.
x is a dummy variable. That's why we can say "replace x with 1/x". It's the same with locus.
But if you don't like saying "replace x with 1/x", you can instead do it like this:
Let y=1/x.
∴ 2/y<sup>3</sup>-5/y<sup>2</sup>+4/y+6=0.
∴ 6y<sup>3</sup>+4y<sup>2</sup>-5y+2=0.
The result is the same, with dummy variable y instead of x.
It's the same in locus problems. In my syllabus solution, instead of saying "replace z with 1/z", you can say "Let w=1/z" and get the same result, i.e, the same circle with the same centre and same radius, but with a dummy variable w instead of z in the equation.
Hope that helps.
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Hey, I don't know if this suffices as a proof or something which is sufficiently within the syllabus, but a possible approach uses the fact that reix = r(cosx + isinx) = rcis(x) (which comes from Euler's formula: http://en.wikipedia.org/wiki/Euler's_formula ... If you look up the taylor series stuff it's understandable using what is within the syllabus and the rest follows).
So that it makes more sense, first consider the simple case where the circle is centered around the origin:
Let f(x) and F(x) be functions where f(x) = reix = rcis[x] = z (and 0 <= x <= 2pi), so f(x) is the anticlockwise circle from 0 to 2pi of radius r in the complex plane, and F(x) = z-1.
F(x) = F(x) = z-1 = (1/r)e-ix which is a circle of radius (1/r) in the complex plane going clockwise from 0 to -2pi.
...
If the circle can be anywhere then let f(x) = reix + z2 = z1 + z2 = z (z2 is any complex number x2 + iy2).
F(x) = z-1
= 1/(z1 + z2)
= conj.(z1 + z2)/(|z1 + z2|)
= (conj.z1 + conj.z2)/k ... (k since the modulus of the denominator will take a real value)
= (r/k)e-ix + (1/k)conj.z2
which is a clockwise cirlce with a radius r/k... note that when z2 = 0 then k = r2 which is the first result. Anyhow, there is my BoS cameo... sorry for any mistakes, I'm quite out of practice.
So that it makes more sense, first consider the simple case where the circle is centered around the origin:
Let f(x) and F(x) be functions where f(x) = reix = rcis[x] = z (and 0 <= x <= 2pi), so f(x) is the anticlockwise circle from 0 to 2pi of radius r in the complex plane, and F(x) = z-1.
F(x) = F(x) = z-1 = (1/r)e-ix which is a circle of radius (1/r) in the complex plane going clockwise from 0 to -2pi.
...
If the circle can be anywhere then let f(x) = reix + z2 = z1 + z2 = z (z2 is any complex number x2 + iy2).
F(x) = z-1
= 1/(z1 + z2)
= conj.(z1 + z2)/(|z1 + z2|)
= (conj.z1 + conj.z2)/k ... (k since the modulus of the denominator will take a real value)
= (r/k)e-ix + (1/k)conj.z2
which is a clockwise cirlce with a radius r/k... note that when z2 = 0 then k = r2 which is the first result. Anyhow, there is my BoS cameo... sorry for any mistakes, I'm quite out of practice.
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conj.(z<sub>1</sub> + z<sub>2</sub>)/(|z<sub>1</sub> + z<sub>2</sub>|)KFunk said:
should be
conj.(z<sub>1</sub> + z<sub>2</sub>)/(|z<sub>1</sub> + z<sub>2</sub>|<sup>2</sup>)
Also, I reckon KFunk'd better write something about the circle not going through (0,0) because otherwise, you get a line, not a circle for the locus of 1/z.
dreampianist needent worry. The syllabus states that linear fractional transformations of circles are not required for the course.
My determinants method proved the locus is a circle without having to find the centre and radius and is the most elegant one. When no_arg prodded me to stick to the syllabus I constructed a more complicated argument finding the centre and radius using methods only in the syllabus. But this is an embrassment. The syllabus methods are grotesque compared to the beauty of the determinants method. So the syllabus committee thought it best to exclude linear fractional transformations of circles altogether. But since when did that stop boredofstudies.org members?
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Unfortunately my real mistake, which I noticed having looked back over it, was in assuming that |reix + z2|2 takes a constant value for all values of x... which is significantly easier to do if you use a z1 substitute like I did. Disregard the second part, the first part for circles centred at (0,0) still holds.
I can see why you prefer a polar form instead of my Cartesian form. But you'd have to show more working and show your centre is conj(z<sub>2</sub>)/(|z<sub>2</sub>|<sup>2</sup>-r<sup>2</sup>) and radius r/||z<sub>2</sub>|<sup>2</sup>-r<sup>2</sup>|.