Given a locus of points, show that it is a circle (find its radius and center)

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SUMMARY

The discussion focuses on demonstrating that the locus of points defined by the transformation {1/z: z ∈ ε L} results in a circle with a specific radius and center. It is established that for the line y=a, the resulting circle has a radius of 1/(2a) and a center at -i/(2a). Additionally, the transformation T(z)=(az+b)/(cz+d) is identified as a linear fractional transformation that maps circles to circles. The mathematical derivation confirms that the locus {1/z: z ∈ ε C} for the circle |z|=r also results in a circle centered at the origin with radius 1/r.

PREREQUISITES
  • Understanding of complex numbers and their representation.
  • Familiarity with linear fractional transformations.
  • Knowledge of circle equations in both rectangular and complex planes.
  • Basic calculus concepts for deriving distances and radii.
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  • Study the properties of linear fractional transformations in detail.
  • Learn how to derive the center and radius of circles in the complex plane.
  • Explore the implications of transformations on geometric shapes in complex analysis.
  • Investigate the relationship between loci of points and their geometric representations.
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Goomba
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1. Suppose L is the line y=a. Show that the locus of points {1/z: z E(epsilon) L} is the circle of radius 1/(2a) with center at -i/(2a).



2. T(z)=(az+b)/(cz+d), ad != bc. T(z) is a linear fractional transformation. Linear fractional transformations take circles to circles, where the class of circles includes regular circles and straight lines.

Equation for circle on the rectangular plane: (x-a)^2+(y-b)^2=r^2 where r is the radius and (a,b) is the center.

Equation for circle on the complex plane: |z-center|=radius, I think??




3. The preceding problem with my solution:
Let C be the circle |z|=r. Show that the locus {1/z: z E(epsilon) C} is another circle with center at the origin. Find its radius.

Let z=a+bi.
Given: |z|=r
r=sqrt(a^2+b^2)

|1/z| = |1/(a+bi)| = |(a-bi)/(a^2+b^2)| = |a/(a^2+b^2) - ib/(a^2+b^2)| = sqrt{[a/(a^2+b^2)]^2 + [-b/(a^2+b^2)]^2} = sqrt[(a^2+b^2)/(a^2+b^2)^2] = sqrt[1/(a^2+b^2)] = 1/r, which is its radius.

I don't know how to show its center as the origin. Or even if I correctly found the radius here. The whole process of finding center and radius given a locus of points confuses me...
 
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Goomba said:
1. Suppose L is the line y=a. Show that the locus of points {1/z: z E(epsilon) L} is the circle of radius 1/(2a) with center at -i/(2a).



2. T(z)=(az+b)/(cz+d), ad != bc. T(z) is a linear fractional transformation. Linear fractional transformations take circles to circles, where the class of circles includes regular circles and straight lines.

Equation for circle on the rectangular plane: (x-a)^2+(y-b)^2=r^2 where r is the radius and (a,b) is the center.

Equation for circle on the complex plane: |z-center|=radius, I think??




3. The preceding problem with my solution:
Let C be the circle |z|=r. Show that the locus {1/z: z E(epsilon) C} is another circle with center at the origin. Find its radius.

Let z=a+bi.
Given: |z|=r
r=sqrt(a^2+b^2)

|1/z| = |1/(a+bi)| = |(a-bi)/(a^2+b^2)| = |a/(a^2+b^2) - ib/(a^2+b^2)| = sqrt{[a/(a^2+b^2)]^2 + [-b/(a^2+b^2)]^2} = sqrt[(a^2+b^2)/(a^2+b^2)^2] = sqrt[1/(a^2+b^2)] = 1/r, which is its radius.

I don't know how to show its center as the origin. Or even if I correctly found the radius here. The whole process of finding center and radius given a locus of points confuses me...
You found the radius by taking |1/z|, that is, the distance from the point 1/z to the origin. In other words, you have shown that the distance from any point to the origin is a constant. Doesn't that show that the locus is a circle with the origin as center?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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