Geometric interpretation of complex equation

[WubbaLubba Dubdub]

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?

 

[ehild]

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?

 

[WubbaLubba Dubdub]

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.

 

[ehild]

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.

 

[Ray Vickson]

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.

 

[haruspex]

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.