Is this true?? (1 Viewer)

no_arg

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Prove that if z ranges over a fixed circle in the complex plane (not containing 0)
then the locus of 1/z is also a circle
 

darkliight

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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.
 

no_arg

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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!)
 
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Mill

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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. :)
 

no_arg

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Anyone have a proof within the syllabus?
What is the radius and centre of the new circle?
 
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Riviet

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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.
 

Riviet

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Raginsheep said:
Isn't that only when mod z is 1?
Yeah you're right, my bad. Already rusty in first week of holidays! :S
 

KFunk

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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.
 
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KFunk

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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.
 

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