# Math Help - Equation of locus of points representing z

1. ## Equation of locus of points representing z

Hi, i got this question and dont fully understand how they solved it:

You have to find the locus of points representing z if:

abs(z-3i)-abs(z+31)=2

my first thought would be to replace z with (x+iy) but what then?

2. Originally Posted by Mahanen
Hi, i got this question and dont fully understand how they solved it:

You have to find the locus of points representing z if:

abs(z-3i)-abs(z+3i)=2

my first thought would be to replace z with (x+iy) but what then?
Then take the absolute value of x + iy - 3i = x + i(y - 3) and x + iy + 3i = x + i(y + 3).

So you have to simplify something of the form $\sqrt{A} - \sqrt{B} = 2 \Rightarrow \sqrt{A} = 2 + \sqrt{B}$.

Square both sides, simplify, square both sides again and simplify. You will get a single branch of a hyperbola (there is an implicit restriction on z you will need to be careful of that. This restriction means that both branches do no get included).

Of course, if you're familiar with the locus definition of a hyperbola then there's a much simpler geometric approach that can be taken.

3. Okay, seems quite simple now. Thanks. This is what i have done:

$\sqrt{x^2+(y-3)^2}-\sqrt{x^2+(y+3)^2}=2$
$\sqrt{x^2+(y-3)^2}=2+\sqrt{x^2+(y+3)^2}$
$x^2+(y-3)^2=4+4\sqrt{x^2+(y+3)^2}+x^2+(y+3)^2$
$x^2+y^2-6y+9=4+4\sqrt{x^2+(y+3)^2}+x^2+y^2+6y+9$
$-12y-4=4\sqrt{x^2+(y+3)^2}$
$-(3y+1)=\sqrt{x^2+(y+3)^2}$
$(3y+1)^2=x^2+(y+3)^2$
$x^2-8y^2+8$
$x^2/8-y^2=1$

And in this case it will only be the bottom branch of the graph.

i have also looked at the locus definition of the hyperbola, didnt think of defining it that way. makes it much simpler just by substituting in:

$\sqrt{(x-c)^2+y^2}-\sqrt{(x+c)^2+y^2}=2a$

into:

$x^2/a^2-y^2/(c^2-a^2)=1$

you get the final equation.