1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Geometric interpretation of complex equation

  1. May 9, 2017 #1
    1. The problem statement, all variables and given/known data
    $$z^2 + z|z| + |z|^2=0$$
    The locus of ##z## represents-
    a) Circle
    b) Ellipse
    c) Pair of Straight Lines
    d) None of these

    2. Relevant equations
    ##z\bar{z} = |z|^2##

    3. The attempt at a solution

    Let ##z = r(cosx + isinx)##
    Using this in the given equation
    ##r^2(cos2x + isin2x) + r^2(cosx + isinx) + r^2 = 0##
    ##r^2(cos2x + cosx + 1 + i(sin2x + sinx)) =0##
    Thus ##r=0## or ##cos2x + cosx + 1 = 0## and ##sin2x + sinx =0##
    I can't think of a geometrical interpretation of this result. Using the real part as x and the imaginary part as y makes a funny graph at wolfram alpha

    But solving analytically, I get a pair of straight lines
    ##z^2 + z\sqrt{z\bar{z}} + z\bar{z}=0##
    ##z(z + \sqrt{z\bar{z}} + \bar{z})=0##
    Thus, either## z = 0## or ##(z + \sqrt{z\bar{z}} + \bar{z})=0##
    In case ##z + \sqrt{z\bar{z}} + \bar{z} =0##
    ##2Re(z) = -\sqrt{z\bar{z}}##
    Let ##z = x + iy##
    ##(\sqrt{3}x - y)(\sqrt{3}x + y)=0##

    Where did I go wrong?
     
  2. jcsd
  3. May 10, 2017 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Do not use double angles. You get cos(x) from the imaginary part 2sin(x)cos(x)+sinx =0, and the same result as you got with the other method.
     
  4. May 10, 2017 #3
    So, when ##cosx = -\frac{1}{2}## the equation is satisfied irrespective of the value of ##r## and so I get a line since r can vary for that particular value of x? Does this make sense? Thanks though, I just didn't know what to do with the equation I had.
     
  5. May 10, 2017 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes. But you can get the value of sinΦ, and then both x=rcos(Φ) and y=rsin(Φ), where r is arbitrary, and you also get the relation y=tan(Φ) x where tan(Φ) is known, and it can take two values.
    In polar coordinates, Φ=constant is a straight line. ##cosx = -\frac{1}{2}## corresponds to two angles, so two straight lines.
     
  6. May 10, 2017 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You did not go wrong; you simply did not complete the analysis.

    However, the given answer possibilities are not really correct: the solutions form a pair of line segments, not a pair of lines. As you have pointed out, the given equation is ##z (z + \bar{z} + |z|) = 0##, so either ##z = 0## or ##2x + \sqrt{x^2+y^2} = 0## (where ##z = x + iy##). In particular, if ##z = x + iy \neq 0## then ##2x = -\sqrt{x^2+y^2} < 0,## so only the portion ##x < 0## is allowed. If append the point ##z = 0##, that means that the allowed solutions are of the form ##(x,y): y = \pm \sqrt{3} x,\: x \leq 0##.

    A couple of final points for future reference:
    (1) Do not use the same symbol ##x## for the real part and also for the argument in the same problem. Typically we denote the argument by ##\theta## or ##\phi## or some other Greek letter, but if you want to avoid excessive typing you can use something like ##w## instead. Just don't use ##x## or ##y##.
    (2) In LaTeX, do not write ##sin x ## or ##cos x##, as these are hard to read and look ugly. Instead, use ##\sin x## and ##\cos x##, obtained by typing "\sin" instead of "sin" and "\cos" instead of "cos". The same applies to other commands like ##\tan, \cot, \sec, \csc, \arcsin, \arccos, \arctan, \sinh, \coth, \tanh, \log, \ln, \lim, \max, \min, ## etc.
     
  7. May 12, 2017 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That's wrong. The |z|2 term has been turned into z|z|. 1+i√3 is a solution of the original equation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Geometric interpretation of complex equation
Loading...