# Geometric interpretation of complex equation

• WubbaLubba Dubdub
In summary: Do not use the same symbol (such as ##z##) for two different objects, such as a complex number and its modulus. That is, let ##|z| = r## or ##|z| = s## or whatever. The same thing applies to the argument, so use ##\theta## or ##\phi## or whatever for that.In summary, the locus consists of two line segments of length ##\sqrt{3}\: r## joined at the origin.
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?

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
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.
WubbaLubba Dubdub said:
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
ehild said:
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.

WubbaLubba Dubdub said:
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.

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.

Ray Vickson said:
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.

## 1. What is the geometric interpretation of a complex equation?

The geometric interpretation of a complex equation involves representing the equation graphically on the complex plane, where the real part of the equation is plotted on the x-axis and the imaginary part is plotted on the y-axis. The solution to the equation is represented by the point where the two axes intersect.

## 2. How does the geometry of a complex equation affect its solutions?

The geometry of a complex equation can greatly affect its solutions. For example, if the equation is a quadratic equation, the geometry of the equation will determine whether it has two distinct solutions, one repeated solution, or no solutions at all.

## 3. Can a complex equation have more than one solution?

Yes, a complex equation can have multiple solutions. This is because on the complex plane, there are infinitely many points that can satisfy the equation. The number of solutions will depend on the degree of the equation and its geometry.

## 4. How do we find the solutions to a complex equation on the complex plane?

To find the solutions to a complex equation on the complex plane, we can use a variety of methods such as graphing, algebraic manipulation, or using the quadratic formula. The solutions will be represented by the points where the equation intersects the x-axis.

## 5. Why is the complex plane useful in understanding complex equations?

The complex plane is a useful tool in understanding complex equations because it provides a visual representation of the equation and its solutions. It allows us to see the relationship between the real and imaginary parts of the equation and how they affect the solutions. Additionally, the complex plane can help us identify patterns and symmetries in complex equations, making them easier to solve.

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