Locus with complex numbers

Therefore, the locus is the two points (-3,0) and (1,0).In summary, the locus given by z(\overline{z}+2)=3, where the overbar means conjugate, is the two points (-3,0) and (1,0) on the x-axis. This can be found by solving for x and y and setting y=0, resulting in the equation (x+1)^2=4.
  • #1
Grand
76
0

Homework Statement


What is the locus given by
[tex]z(\overline{z}+2)=3[/tex]

where the overbar means conjugate.

Homework Equations


The Attempt at a Solution


After using z=x + yi and expanding the backets, one gets the equation:
[tex]x^2+2x+y^2+2iy=3[/tex]
or
[tex](x+1)^2+y^2 +2iy=4[/tex]

which is a circle crossed with the line y=0, which means that the locus is actually the points +/-1. However, the book says it is a circle.
 
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  • #2
You're correct that y=0, but you didn't solve for x correctly.
 
  • #3
Oh yes, sorry for that, it must be 1 and -3 (if I'm not super absent-minded again).
 
  • #4
I think that u can't just times it in the complex number when u want to find the locus.

My teacher taught me that when u want to find a locus of an equation u have to find the magnitude of unit vector of it.

Example : From question, (x+yi)(x-yi+2)=3

The magnitude unit vector of x+yi is (x^2+y^2)^1/2
The magnitude unit vector of x+2-yi is ((x+2)^2+(-1)^2)^1/2

Then we times it up and we get (x^2+y^2)^1/2 x ((x+2)^2+(-1)^2)^1/2 = 3

the result will like be x^4+4x^3+4x^2+2(xy)^2+4xy^2+4y^2+y^4=9

Ya i think i misunderstood the qeustion

I can see that it's not a circle from the equation (correct me if i am wrong)

Ya i think i misunderstood the qeustion..it is not absolude l l
 
Last edited:
  • #5
Grand said:
After using z=x + yi and expanding the backets, one gets the equation:
[tex]x^2+2x+y^2+2iy=3[/tex]
or
[tex](x+1)^2+y^2 +2iy=4[/tex]

To make it even more obvious to the reader/marker that you know what is happening, you might want to factorize y as you did with x by noting that [tex](y+i)^2=y^2+2iy-1[/tex]
 
  • #6
I was thinking about that, but if I write it as
[tex](x+1)^2+(y+i)^2=3[/tex]

it is clear that it is a circle. So which one is actually true?
 
  • #7
How is that a circle? The presence of i screws it up.
 
  • #8
Your original formula, [itex]z(\overline{z}+ 2)= 3[/itex] educes, after you write z= x+ iy, to [itex](x+ 1)^2+ (y+ i)^2= 3[/itex] but that is NOT a circle in the xy-plane because, having change to [itex]x+ iy[/itex], both x and y must be real numbers. There is no point "(-1, -i)" in the xy-plane.

[itex]z(\overline{z}+ 2)= 3[/itex] gives, just as you say, [itex](x+1)^2+ y^2+ 2iy= 3[/itex] and comparing real and imaginary parts we have [itex](x+1)^2+ y^2= 3[/itex] and 2y= 0. Yes, y= 0 and then we have [itex](x+ 1)^2= 3[/itex] so that [itex]x= -1\pm\sqrt{3}[/itex]. The locus is the two points [itex](-1+ \sqrt{3}, 0)[/itex] and [itex](-1- \sqrt{3}, 0)[/itex], the only two complex numbers that satisfy the original equation.
 
  • #9
Oh yes sorry, I screwed up big time.
 
  • #10
so actually mentallic, was my method correct or wrong?
 
  • #11
No not quite. As Hallsofivy has already shown, while the solutions lie on that implicit equation you've given (which I'm sure is an ellipse by the way), they only exist for when y=0, or in other words where it cuts the x-axis.
By the way, small typo on Hallsofivy's part, the solutions are (-3,0) and (1,0).
 
  • #12
so what is the equation of the locus?
 
  • #13
The intersection between the equations [tex]y=0[/tex] and [tex](x+1)^2+y^2=4[/tex]
 

What is a locus with complex numbers?

A locus with complex numbers is a set of points in a complex plane that satisfy a given condition or equation. It represents the solution set to a complex equation.

How is a locus with complex numbers different from a locus with real numbers?

A locus with complex numbers involves the use of imaginary numbers, while a locus with real numbers only involves real numbers. This means that the points in a locus with complex numbers can have a real and imaginary value, while the points in a locus with real numbers can only have a real value.

What are some common shapes that can be represented by a locus with complex numbers?

Some common shapes that can be represented by a locus with complex numbers include lines, circles, ellipses, and parabolas. These shapes can have both real and imaginary parts to their equations, allowing for more complex and interesting curves.

How are loci with complex numbers used in science?

Loci with complex numbers are used in many fields of science, including physics, engineering, and mathematics. They are often used to model and solve complex systems and equations, as well as to represent physical phenomena such as electrical circuits and quantum mechanics.

What are some real-world applications of loci with complex numbers?

Some real-world applications of loci with complex numbers include designing electrical circuits, analyzing the behavior of waves, and studying the motion of objects in space. They are also used in computer graphics to create complex and realistic visual effects.

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