Find locus of a complex number (1 Viewer)

[schmeichung]

arg(z-1)-arg(z+1)=- pi/3
how do I get the locus equation?

Thanks

 

[steverulz55]

the locus can be interpret geomtrically:

you can probably try algebraically:

arg [ (z - 1) / (z + 1) ] = -pi/3
let z = x+iy etc, but thats probably too messy and long

not if it is arg (z-1) - arg (z+1) = pi/3 , the locus is in the second and first quadrants

 

[schmeichung]

thanks
but I mean the locus (the circle) equation with limit
I dont know how to find the circle equation

 

[schmeichung]

I know the equation can be obtained by using circle geometry property (the angle is twice at the centre of the angle to the subtended to circumference from same segment) something like that
but I am not sure what to do

 

[turtle_2468]

Draw an argand diagram. Note the points 1 and -1, and draw z anywhere (for the time being), such that you do get an angle between 1) the line joining 1 and z
2) the line joining -1 and z
around pi/3. Now if you drew z above the y axis, you'll see that arg(z-1)>arg(z+1) and that's bad because we want -pi/3... so draw z at around the same point below the y axis and you'll get -pi/3. Remember that if you have something of this form, the locus will be the arc of a circle (angles bounded by equal arcs... remember?) This circle will have points 1, -1, and one more point is sufficient to determine where the circle is. Now remember that an equilateral triangle has angles pi/3, so construct the equilateral triangle down from the line joining -1 and 1, getting the other point (0, -root(3)/2). So... letting the circle containing those three points be C, the locus of the number is:

The arc of C that includes (0, -root(3)/2) (not including -1 and 1) (edit: I mean -root(3)i/2, but I hope you understand)

I don't have paper right now, but I think that'd be something like... umm... |z+(i/(2root(3))|^2= 4/3? Something like that..

 

[turtle_2468]

I know the equation can be obtained by using circle geometry property (the angle is twice at the centre of the angle to the subtended to circumference from same segment) something like that
but I am not sure what to do

In reply to what you said before...
Let the intersection of the arc with the y-axis be Y.
Then Y, -1, 1 form an equilateral triangle, you can work out Y from that, and the centre of the triangle is the same as the centre of the circle..

 

[ngai]

thanks
but I mean the locus (the circle) equation with limit
I dont know how to find the circle equation

clearly centre is on the imaginary axis
so draw the centre in there somewhere, and join it to 1 and -1
then that angle at centre = 2(angle at circumference) = 120
so u have two triangles, with angles 60 at the centre
and both triangles have a rt angle at the origin
so theyre 30 60 90 triangles
and then u can find the centre, which is -i/rt3
and radius not hard to find..pythagoras will do

hope u understand all that...coz i cbb scanning in a diagram

 

[ngai]

In reply to what you said before...
Let the intersection of the arc with the y-axis be Y.
Then Y, -1, 1 form an equilateral triangle, you can work out Y from that, and the centre of the triangle is the same as the centre of the circle..

haha whoops didn't read that