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Locus complex plane

  1. Jul 12, 2005 #1
    Q. For the real constant a find the loci of all points z = x + yi in the complex plane that satisfy:

    a) [tex]{\mathop{\rm Re}\nolimits} \left\{ {\log \left( {\frac{{z - ia}}{{z + ia}}} \right)} \right\} = c,c > 0[/tex]

    b) [tex]{\mathop{\rm Im}\nolimits} \left\{ {\log \left( {\frac{{z - ia}}{{z + ia}}} \right)} \right\} = k,0 \le k \le \frac{\pi }{2}[/tex]

    I have very little idea as to how to do these questions.

    For each of them I've thought about first 'ignoring' the Im and Re to see where I could get. I thought, maybe exponentiate both sides but then I'm still left with a quotient of involving z, with the quotient being equal to the exponential of a positive number or an angle(depending on if I'm working on part a or b). There doesn't seem to be an easy way to do this question.

    I get the feeling that perhaps these two require some sort of geometric interpretation but I can't really see anyway to interpret the equation. Can someone please help me get started on deducing what the locus of points for each question is?
  2. jcsd
  3. Jul 12, 2005 #2
    The complex logarithm has a specific definition involving a natural log and an argument angle. That is, it's slightly different then the one that you may be use to when dealing with just real numbers. Find this definition and start there.
  4. Jul 12, 2005 #3
    Do you mean where log(z) = log|z| + iargz? I'll look it up again anyway, I recently used it. Thanks for the help.
  5. Jul 14, 2005 #4


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    There's no "easy" (meaning elegant, short) way to do this question that I know of, but you can slog through the algebra fairly directly.

    Just express z = x + yi and arrange the numerator and denominator in Cartesian form. Then use the definition of log of a complex number (as you've already stated), and separate into real and imaginary parts.

    Immediately, you'll find that both loci are obvious conic sections. On closer inspection, they're both circles. Find the center and radii in each case (if you need to visualise/sketch the locus), and you're golden. :smile:

    The working is too long for me to reproduce (and I'm still feeling under the weather), but here are the answers I got :

    Part a) (the Real part) :

    [tex]x^2 - [y - a(\frac{1 + e^{2c}}{1 - e^{2c}})]^2 = 4a^2\frac{e^{2c}}{(1-e^{2c})^2}}[/tex]

    Part b) (the Imaginary part) :

    [tex]{(x + a\cot k)}^2 + y^2 = a^2\csc^2{k}[/tex]

    Compare each of the above to the general circle [tex]{(x - A)}^2 + {(y - B)}^2 = R^2[/tex], which is centered at (A,B) and has radius R in order to see what the center and radius are in each case.
    Last edited: Jul 14, 2005
  6. Jul 15, 2005 #5
    Thanks for those answers Curious3141, I'll see what answer I can come up with. I haven't gotten around to trying to do it the long(ie. algebraic way) yet.

    Edit: Looking at your second answer I'm not sure how to interpret the cosec(k) part because k takes on values between 0 and pi/2. So does that have any affect on the radius? I'm finding it a little difficult to comprehend a circle with a varying 'radius.' Dividing both sides by the RHS I would get an ellipse I think by then the smi-minor and semi-major axes would vary. :confused:
    Last edited: Jul 15, 2005
  7. Jul 15, 2005 #6


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    For a given case, k takes on a single value within a restricted range. It doesn't vary.

    I should've specified that for k = 0 (and nonzero a), you don't get a circle (or any curve), just a straight vertical line at [itex]x = 0[/itex]. You can determine this by letting k tend to zero in my expression and seeing what happens. You'll need L'Hopital's Rule to reduce a limit. Personally, I feel the restriction should've been a half-closed interval (open at the lower limit), but it's no big deal.

    BTW, a (non-circular) ellipse only results when the coefficients of [itex]x^2[/itex] and [itex]y^2[/itex] are different after you put the equation into a standard form. This is definitely a perfect circle for most values of k and all nonzero values of a.
    Last edited: Jul 15, 2005
  8. Jul 15, 2005 #7
    I think I see what you mean now, thanks again for the help.
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