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Homework Help: Complex Circle Equation with random variable attached to Z.

  1. Mar 23, 2014 #1
    1. The problem statement, all variables and given/known data
    |zi - 3| = Pi

    2. Relevant equations
    Well, it clearly has to do with a circle but I do not believe there is a general equation for what I am asking about.

    3. The attempt at a solution
    There is no general solution not trying to solve anything.

    I want to know exactly what happens if there is a variable attached to z in a complex circle equation.

    For example, |z - 3| = Pi is a circle centered at 3,0 with radius pi on the complex plane.
    However, I do not quite understand what I should do if there is a variable such as i attached to z in the equation.

    I do not wish to be picky, but I'm not looking for any secret hints just a complete answer on what happens if there is a variable attached to z in the equation.
  2. jcsd
  3. Mar 23, 2014 #2
    Sorry, we don't give out complete answers, only secret hints. PF Rules: https://www.physicsforums.com/showthread.php?t=414380
  4. Mar 23, 2014 #3
    Yes I do understand that but how could you give a secret hint to such a question? I can understand hinting someone towards an answer but this type of question you really cant hint towards an answer too, well at least not in my experience as a student.

    "On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made."

    While I support this, and I do realize that this is posted under homework. I just did not recognize a better place to post it. This is a general question about how complex circle equations work. It is like I am asking how a hyperbola reacts when you do this or that to said equation.

    In a sense it is asking how a concept works, something that is extremely hard to hint or provide advice on without actually going into some sort of proof / direct answer.
    Last edited: Mar 23, 2014
  5. Mar 23, 2014 #4

    Ray Vickson

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    Try to be more specific: I do not understand what you mean by "attaching" something to z in the circle. Are you saying (for example) that you have a single real-valued random variable ##X## with, say, a known probability density function ##f(\cdot)## and you then construct a new type of random object over the complex plane, by something like ##Z = x + X + i y##; this attaches ##X## to the point ##z = x + iy## by centering it at ##z## and letting it stretch along a line through ##z## parallel to the real axis. Or, do you have something else in mind? Not being a mind-reader, I cannot really figure out what you want.
  6. Mar 23, 2014 #5
    Sorry about that. I'll try to be exact as possible.

    In the equation |z*i - 3| = Pi, there is the imaginary complex number i attached to z that I cannot seem to fathom. If it were just |z - 3| = Pi, that would just be a circle no problem.

    The logical step that I took was to expand it into |(X+IY)*I - 3| = PI and then turn that into the

    Squareroot of (X^2) + (-Y-3)^2 = R^2 = PI. This would give me a circle with center of (0,-3) and radius PI. This is where I stopped since I was not 100% certain this is how I should handle these kind of problems.
  7. Mar 23, 2014 #6


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    I think this is a really good question! Here are some hints:

    What happens when you multiply a complex number by i? Can you describe this geometrically?


    If |zi - 3| = r, then can you see a formula for |z - something| = r?
  8. Mar 23, 2014 #7

    Ray Vickson

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    ##iz - 3 = i(z + 3i)##, so ##|iz - 3| = |i| |z+3i| = |z+3i|= |x + i(3+y)|##.
  9. Mar 23, 2014 #8
    I also tried to think about it like this but hit some brick walls.

    When I multiply for say Z = X + IY by I, it would switch the places of X and Y and make Y negative.
    I could not fathom how that would work out on the complex plane though, I mean I guess you could say the axis of X and Y may switch place but that seems really farfetched and needs confirmation. Sadly I couldn't describe it any further than that.

    For the second part, If I wanted specifically that form, I could divide the entire thing (not sure if I can) by I, but then I would have a radius that would be mixed with I and I can't fathom an imaginary radius either.
  10. Mar 23, 2014 #9
    Is it safe to assume that when you factor out the absolute I, it will always disappear?
  11. Mar 23, 2014 #10


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    Keep going with this. Plot some points and see where they move to when you multiply by i. Start with (1, 0), (1, 1), (0, 1), (-1, 1), (-1, 0), (-1, -1), (0, -1), (1, -1)

    Map where these move to and see whether you can spot a pattern.
  12. Mar 23, 2014 #11
    No it doesn't disappear, ##|i|=1##. Can you see it now?
  13. Mar 23, 2014 #12

    Ray Vickson

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    Yes. For any two complex numbers ##z_1,z_2## we have, in general, ##|z_1 z_2| = |z_1| \, |z_2|##.

    So, if ##u,v,z## are all complex with (##u \neq 0##), we have
    [tex] |uz + v| = \left|u\left(z + \frac{v}{u} \right) \right| = |u| \, |z + w|[/tex]
    where ##w = v/u##. So, if ##r > 0## the equation ##|uz + v| = r## is the same as ##|z + w| = r/|u|##; this is a circle of radius ##r/|u|##, centered at ##-w##.
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