Find Equation of Locus of z for \arg(\frac{z-2i}{z-3}) = \frac{\pi}{6}

[kidsmoker]

Homework Statement

Find the equation of the locus of z where

$$<br /> \arg(\frac{z-2i}{z-3}) = \frac{\pi}{6}<br />$$

Homework Equations

$$<br /> \arg(x+iy) = \arctan(\frac{y}{x})<br />$$

The Attempt at a Solution

I wrote z=x+iy in which case we have

$$\frac{z-2i}{z-3} = \frac{x+iy-2i}{x-3+iy} = \frac{(x^3 - 3x + y^2 - 2y)+(6 - 2x - 3y)}{(x-3)^2+y^2}$$

I'm ommiting my working obviously, but I've checked it and i think it's correct. This means that

$$<br /> \arctan(\frac{6-2x-3y}{x^2 -3x + y^2 - 2y}) = \frac{\pi}{6}<br />$$

$$<br /> 18 - 6x - 9y = \sqrt{3} (x^2 - 3x + y^2 - 2y)<br />$$

$$<br /> (x + \frac{2\sqrt{3} - 3}{2})^2 + (y + \frac{3\sqrt{3} - 2}{2})^2 = 13<br />$$

which is the equation of a circle. This seems correct, but I think the locus is only part of the circle not the whole thing. How would you guys go about tackling this problem?

Thanks.

 

[tiny-tim]

… which is the equation of a circle. This seems correct, but I think the locus is only part of the circle not the whole thing. How would you guys go about tackling this problem?

Hi kidsmoker!

Personally, I'd use ordinary geometry.

Let A = 2i, B = 3, then the equation is angle BXA = 30º (taking care not to get -30º or 390º).

And the locus of points X with BXA constant is … ?

And just check "by hand" the solutions you have to ignore because you get 390º.

 

[kidsmoker]

Okay so X lies on an arc of a circle passing through A and B (above to the right)?

 

[tiny-tim]

Okay so X lies on an arc of a circle passing through A and B (above to the right)?

Yes!

(erm … forget what I said about 390º … I forgot exactly what the question was. )

 

[kidsmoker]

Cool thanks.

Yeah I was wondering what all that 390º stuff was about lol.