# Homework Help: Locus with complex numbers

1. Sep 16, 2010

### Grand

1. The problem statement, all variables and given/known data
What is the locus given by
$$z(\overline{z}+2)=3$$

where the overbar means conjugate.

2. Relevant equations

3. 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.

2. Sep 16, 2010

### vela

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

3. Sep 16, 2010

### Grand

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

4. Sep 16, 2010

### tee yeh hun

I think that u cant 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: Sep 17, 2010
5. Sep 16, 2010

### Mentallic

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

6. Sep 16, 2010

### 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?

7. Sep 16, 2010

### vela

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

8. Sep 16, 2010

### 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.

9. Sep 16, 2010

### Mentallic

Oh yes sorry, I screwed up big time.

10. Sep 16, 2010

### tee yeh hun

so actually mentallic, was my method correct or wrong?

11. Sep 16, 2010

### 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).

12. Sep 17, 2010

### tee yeh hun

so what is the equation of the locus?

13. Sep 17, 2010

### Mentallic

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