WubbaLubba Dubdub said:
Homework Statement
$$z^2 + z|z| + |z|^2=0$$
The locus of ##z## represents-
a) Circle
b) Ellipse
c) Pair of Straight Lines
d) None of these
Homework Equations
##z\bar{z} = |z|^2##
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?
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.
