Least distance between two complex numbers on two loci

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SUMMARY

The discussion revolves around solving a CIE A'level maths P3 problem from the October/November 2013 exam, specifically question 8.(b), which involves finding the least distance between two complex numbers on specified loci. The first locus is a circle centered at (0,-1) with a radius of 1, and the second is a line at 135 degrees to the horizontal starting at x=2. Participants confirm the solution simplifies to √2 + 1/√2 - 1 and discuss methods for minimizing the distance |z-w|, including calculus-based approaches and geometric methods.

PREREQUISITES
  • Understanding of complex numbers and their representations
  • Knowledge of locus definitions, specifically circles and lines
  • Familiarity with calculus concepts, particularly derivatives
  • Ability to apply geometric transformations in problem-solving
NEXT STEPS
  • Study the calculus-based method for minimizing distance in complex analysis
  • Learn about geometric transformations and their applications in locus problems
  • Explore the implications of rotating points in the complex plane
  • Investigate additional complex number problems involving loci and distances
USEFUL FOR

Students preparing for A'level mathematics, educators teaching complex number geometry, and anyone interested in advanced problem-solving techniques in mathematics.

moriheru
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Homework Statement


This is a CIE A'level maths P3 question out of an exam from 2013 in October/November. As there is no markscheme ( I at least can't find one), I would be grateful if someone could look at my solution to the problem and correct me if I made a mistake.
The problem is 8.(b) below.
IMG_1880.PNG

Homework Equations

The Attempt at a Solution


The first locus they are asking for is that of a circle with centre (0,-1) and radius 1 and the second locus is a line 135 deg. to the horizontal (real number axis) starting at x=2. I call z1 and z2 the points which will give the least value of abs(z-w). Both these points must lie on a line l2. My further working and sketch of loci and the line are in the following image.In the last step I use the distance formula for the two complex numbers I calculated in the earlier steps. In the earlier steps I equated the equation of l2 and the equations for the loci.
IMG_1908.JPG


Thanks for any effort! And sorry for the clumsy exposition!
 
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moriheru said:

Homework Statement


This is a CIE A'level maths P3 question out of an exam from 2013 in October/November. As there is no markscheme ( I at least can't find one), I would be grateful if someone could look at my solution to the problem and correct me if I made a mistake.
The problem is 8.(b) below.
View attachment 195845

Homework Equations

The Attempt at a Solution


The first locus they are asking for is that of a circle with centre (0,-1) and radius 1 and the second locus is a line 135 deg. to the horizontal (real number axis) starting at x=2. I call z1 and z2 the points which will give the least value of abs(z-w). Both these points must lie on a line l2. My further working and sketch of loci and the line are in the following image.In the last step I use the distance formula for the two complex numbers I calculated in the earlier steps. In the earlier steps I equated the equation of l2 and the equations for the loci.
View attachment 195848

Thanks for any effort! And sorry for the clumsy exposition!
I get the same result. You can simplify it to √2 + 1/√2 - 1
 
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Thanks! That is great! Is my method correct and is there any general method how to approach these problems geometrically or is it just case by case observation? I know of a calculus based approach. As far as I know one expresses abs(z-w) in terms the general coordinates of any point on the loci and then sets the derivative equal to zero. Is this correct? And how would the general form for abs(z-w) look like? Thanks for any further effort?
 
moriheru said:
is there any general method how to approach these problems geometrically or is it just case by case observation?
A general method (not just circles and lines) will necessarily be by calculus. Of course, minimising |z-w| is the same as minimising |z-w|2, which simplifies things a little.
In many cases, it will be a bit easier with a geometric approach. In this one, I did it by rotating the circle's centre through 45 degrees about the origin. Then I only needed the horizontal distance from the circle to the line x=√2.
 
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Thanks a lot ! And interesting approach. I will keep that in mind.
 

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