Complex numbers / cartesian equations etc

  • Thread starter dj_silver
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  • #1
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Main Question or Discussion Point

Using converse of alternate segment theorem (i think it is)

i.e. this:
"If the line joining two points A and B subtends equal magnitude angles at two other points on the same side of it, then the four points lie on a circle"

establish the cartesian equation, range and domain of the locus of points:

Arg(z - a) - Arg(z) = pi/2

a = 1 + 0i

and for

Arg(z) - Arg(z-a) = pi/2

a = 1 + 0i

say z = x + iy ( i think )

i get

( x - 1/2 ) ^2 + y^2 = 1/4

But i'm not sure about the domain, range etc

If anyone could help me, or point me in the right direction, that would be great.

Thanks.
 

Answers and Replies

  • #2
AKG
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I can't really tell what it is you're trying to do at all.
 
  • #3
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I've got two complex numbers,

z and a
Let z = x + iy and a = 1 + 0i

And I want to establish a cartesian equation using this theorem:

"If the line joining two points A and B subtends equal magnitude angles at two other points on the same side of it, then the four points lie on a circle"

Basically, I'd like to determine the equation, range and domain of the locus of points such that

Arg (z-a) - Arg(z) = pi/2

Sorry if i'm not being clear, i'm finding it hard to explain :p
 
  • #4
AKG
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I don't see how the top and bottom relate. Do you want the cartesian equation of a circle (the locus of points I assume you're talking about)? Do you want to express your equation (with the "Arg"s) in Cartesian form?

Anyways, I'm not sure exactly what you want, but if we have z = x + iy, then arg(z) = arctan(y/x). So:

[tex]\arg (z - a) - \arg (z) = \pi /2[/tex]

[tex]\arctan \left (\frac{y}{x - 1} \right ) - \arctan \left (\frac{y}{x} \right ) = \pi /2[/tex]

[tex]\tan \left [ \arctan \left (\frac{y}{x - 1} \right ) - \arctan \left (\frac{y}{x} \right ) \right ] = \tan \frac{\pi}{2}[/tex]

[tex]\frac{\frac{y}{x - 1} - \frac{y}{x}}{1 + \frac{y}{x - 1}\frac{y}{x}} = \tan \frac{\pi}{2}[/tex]

[tex]\frac{y}{x(x - 1) + y^2} \times \frac{x(x - 1)}{x(x - 1)} = \tan \frac{\pi}{2}[/tex]

Since I'm not exactly sure what you want, I won't go further.
 
  • #5
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I used vectors and scalar product for the first one, I thought it might be relevant to this.

I'm after the cartesian of a circle - the locus of points. (I don't need the "Arg"s in there :))

thanks
 
  • #6
AKG
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Your confusing me even further. Scalar product of what? The cartesian equation of a circle is quite simply:

x² + y² - r² = 0

Of course, if it has its center elsewhere, you'd have to find the center and modify the equation. Another thing you could do:

|z| - r² = 0
 
  • #7
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don't worry, i did it myself

it's (x - 1/2)^2 + y^2 = 1/4

with a domain of (0, 1)
and a range of (0, 1/2]

Scalar product of vectors.

Complex numbers can be represented as vectors, and then i used the scalar (dot) product.

Thanks anyway
 
  • #8
AKG
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dj_silver said:
don't worry, i did it myself

it's (x - 1/2)^2 + y^2 = 1/4

with a domain of (0, 1)
and a range of (0, 1/2]
Yes, that's standard for any circle, although the domain is [0,1] and the range is [-1/2, 1/2]. A circle centered at (a,b) with radius r has the cartesian equation:

[tex](x-a)^2 + (y-b)^2 - r^2 = 0[/tex]
Scalar product of vectors.

Complex numbers can be represented as vectors, and then i used the scalar (dot) product.

Thanks anyway
I know what a scalar product and complex numbers are, I just didn't know which specific vectors/complex numbers you were talking about. This is odd, it seems you're learning about vectors and complex numbers before learning basic stuff about circles? Anyways, I suppose there's nothing wrong with that.
 
  • #9
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It's the domain for which the locus of points exists.... not the circle

for any z below the x axis, Arg(z-a) - Arg(z) is not pi/2, it's -pi/2

(Also z = 0 + 0i gives Arg(z) as undefined - this is the reason for excluding (0,0))

Capische?
 

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