• Support PF! Buy your school textbooks, materials and every day products Here!

Geometric interpretation of complex equation

  • #1

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?
 

Answers and Replies

  • #2
ehild
Homework Helper
15,408
1,810

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
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.
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?
 
  • #3
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.
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.
 
  • #4
ehild
Homework Helper
15,408
1,810
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.
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.
 
  • #5
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

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.
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,532
4,974
the given equation is ##z (z + \bar{z} + |z|) = 0##,
That's wrong. The |z|2 term has been turned into z|z|. 1+i√3 is a solution of the original equation.
 

Related Threads for: Geometric interpretation of complex equation

Replies
3
Views
1K
Replies
13
Views
5K
Replies
26
Views
1K
Replies
2
Views
786
Replies
2
Views
1K
  • Last Post
Replies
1
Views
623
Replies
3
Views
6K
  • Last Post
Replies
3
Views
854
Top