What Is the Correct Locus for the Equation z(overline{z}+2)=3?

[Grand]

Homework Statement

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

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:
x^2+2x+y^2+2iy=3
or
(x+1)^2+y^2 +2iy=4

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.

 

[vela]

You're correct that y=0, but you didn't solve for x correctly.

 

[Grand]

Oh yes, sorry for that, it must be 1 and -3 (if I'm not super absent-minded again).

 

[tee yeh hun]

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

 

[Mentallic]

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

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 (y+i)^2=y^2+2iy-1

 

[Grand]

I was thinking about that, but if I write it as
(x+1)^2+(y+i)^2=3

it is clear that it is a circle. So which one is actually true?

 

[vela]

How is that a circle? The presence of i screws it up.

 

[HallsofIvy]

Your original formula, z(\overline{z}+ 2)= 3 educes, after you write z= x+ iy, to (x+ 1)^2+ (y+ i)^2= 3 but that is NOT a circle in the xy-plane because, having change to x+ iy, both x and y must be real numbers. There is no point "(-1, -i)" in the xy-plane.

z(\overline{z}+ 2)= 3 gives, just as you say, (x+1)^2+ y^2+ 2iy= 3 and comparing real and imaginary parts we have (x+1)^2+ y^2= 3 and 2y= 0. Yes, y= 0 and then we have (x+ 1)^2= 3 so that x= -1\pm\sqrt{3}. The locus is the two points (-1+ \sqrt{3}, 0) and (-1- \sqrt{3}, 0), the only two complex numbers that satisfy the original equation.

 

[Mentallic]

Oh yes sorry, I screwed up big time.

 

[tee yeh hun]

so actually mentallic, was my method correct or wrong?

 

[Mentallic]

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).

 

[tee yeh hun]

so what is the equation of the locus?

 

[Mentallic]

The intersection between the equations y=0 and (x+1)^2+y^2=4