Complex Numbers

  • Thread starter kidsmoker
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  • #1
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Homework Statement



Find the equation of the locus of z where

[tex]

\arg(\frac{z-2i}{z-3}) = \frac{\pi}{6}

[/tex]


Homework Equations



[tex]

\arg(x+iy) = \arctan(\frac{y}{x})

[/tex]

The Attempt at a Solution



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

[tex]\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}[/tex]

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

[tex]

\arctan(\frac{6-2x-3y}{x^2 -3x + y^2 - 2y}) = \frac{\pi}{6}

[/tex]

[tex]

18 - 6x - 9y = \sqrt{3} (x^2 - 3x + y^2 - 2y)

[/tex]

[tex]

(x + \frac{2\sqrt{3} - 3}{2})^2 + (y + \frac{3\sqrt{3} - 2}{2})^2 = 13

[/tex]

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.
 
Last edited:

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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… 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! :smile:

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 … ? :smile:

And just check "by hand" the solutions you have to ignore because you get 390º. :smile:
 
  • #3
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Okay so X lies on an arc of a circle passing through A and B (above to the right)?
 
  • #4
tiny-tim
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Okay so X lies on an arc of a circle passing through A and B (above to the right)?

Yes! :smile:

(erm … forget what I said about 390º … I forgot exactly what the question was. :redface:)
 
  • #5
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Cool thanks.

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

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