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Complex loci problem

  1. Aug 20, 2007 #1
    1. The problem statement, all variables and given/known data

    Given that a complex number, [tex]w[/tex], is such that [tex]\pi / 2< arg(z) < \pi[/tex],

    Sketch the locus described by [tex] arg(z+w^*)=arg(-iw)[/tex]

    2. Relevant equations

    3. The attempt at a solution

    Okay, so I attempted this question and got completely confused. I am thinking that the answer is to shade the first and fourth quadrant completely, that's an intuition (although there can be an explanation for this intuition, but I have no idea how to sketch it out online, sorry). But generally, how should I go about attempting this question? Over here, [tex]w[/tex] is defined by a range instead of any distinct line, thus I am getting the problem of carrying this definition over to my solution of the problem.
  2. jcsd
  3. Aug 25, 2007 #2
    By the constraints of the problem, the locus exists in the second quadrant such that the angle it makes at any point with the x axis is equal to the angle the complex number -iw makes with the x axis. If w is inclined at an angle [tex]\theta[/tex], then -iw is inclined to the x axis at an angle [tex]-(\theta +\frac{\pi}{2})[/tex].
  4. Aug 26, 2007 #3
    Yes, so what I was thinking is that is the answer "shade the entire first and fourth quadrant?"
  5. Aug 26, 2007 #4
    Not the fourth quadrant, because the argument is restricted from pi/2 to pi. Whoa... I wrote a whole post thats missing...basically, it came out to be a straight line segment with the slope tan(3pi/2+theta). y=tan(3pi/2+theta)x is what finally came out. This has to be restricted to the second quadrant, hence its a line segment and not a whole line. Use the eular form of i to get the angle, and since its -(theta+pi/2), the final angle is tan(pi-(-(theta+pi/2))).
  6. Aug 27, 2007 #5
    Why is it only a line segment instead of an area? Theta takes values between pi and pi/2 isn't it?
  7. Aug 30, 2007 #6
    Its a line segment because the equation describes a certain family of lines with a given slope. The position of the line may vary, but that doesnt make it an area.
  8. Aug 31, 2007 #7
    Doesn't the varying of the line make it an area? As in, how do you describe the locus by drawing?
  9. Aug 31, 2007 #8
    are you sure its [tex]-(\theta +\frac{\pi}{2})[/tex] and not [tex]\theta -\frac{\pi}{2}[/tex]?
  10. Aug 31, 2007 #9
    Yeah. Im sure. There's a minus sign to consider.
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