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Math Help - [SOLVED] Three colinear complex points

  1. #1
    Forum Admin topsquark's Avatar
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    [SOLVED] Three colinear complex points

    Hello, this question is so much a "how to do it" as it is a "why did they do it that way?"

    The problem is to find a condition on three complex numbers z_1, ~z_2, ~z_3 showing that they are colinear in the Argand plain.

    Obviously the condition is going to be that the slope through any two sets of points must be equal. In the work that I did I used the slope between points z_1, ~z_3 and between z_2, ~z_3.

    The book, however, went in a very screwy direction to my thinking. Their result is that the quantity \frac{z_1 - z_3}{z_2 - z_3} must be real for the points to be colinear. As it happens this condition is exactly the same as mine, just written in a much neater form.

    But how the would you go about doing this problem and say "Oh yeah! I'd get this..." Is there any significance to the fraction \frac{z_1 - z_3}{z_2 - z_3} that I'm not seeing? Why might the book have put the answer in this form?

    Thanks!

    -Dan
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by topsquark View Post
    Hello, this question is so much a "how to do it" as it is a "why did they do it that way?"

    The problem is to find a condition on three complex numbers z_1, ~z_2, ~z_3 showing that they are colinear in the Argand plain.

    Obviously the condition is going to be that the slope through any two sets of points must be equal. In the work that I did I used the slope between points z_1, ~z_3 and between z_2, ~z_3.

    The book, however, went in a very screwy direction to my thinking. Their result is that the quantity \frac{z_1 - z_3}{z_2 - z_3} must be real for the points to be colinear. As it happens this condition is exactly the same as mine, just written in a much neater form.

    But how the would you go about doing this problem and say "Oh yeah! I'd get this..." Is there any significance to the fraction \frac{z_1 - z_3}{z_2 - z_3} that I'm not seeing? Why might the book have put the answer in this form?

    Thanks!

    -Dan
    Think of them as point in \mathbb{R}^2, then z_1-z_3 is the vector from z_3 to z_1,
    and z_2-z_3 is the vector from z_3 to z_2. Then the points are colinear if
    and only if z_1-z_3 is a (real) scalar multiple of z_2-z_3, but this is equivalent to:


    {\rm{Img}} \left(\frac{z_1 - z_3}{z_2 - z_3}\right)=0

    RonL
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Think of them as point in \mathbb{R}^2, then z_1-z_3 is the vector from z_3 to z_1,
    and z_2-z_3 is the vector from z_3 to z_2. Then the points are colinear if
    and only if z_1-z_3 is a (real) scalar multiple of z_2-z_3, but this is equivalent to:


    {\rm{Img}} \left(\frac{z_1 - z_3}{z_2 - z_3}\right)=0

    RonL
    (sigh) I'm just not grokking this. Am I missing the obvious somewhere in here? I just don't see how the two vectors being scalar multiples of each other translates into \text{Img} \left(\frac{z_1 - z_3}{z_2 - z_3}\right)=0

    -Dan
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    This may not be different way of looking at the problem.
    The line determined by z_1 \,\& \, z_2 is l(t) = z_1  + t\left( {z_2  - z_1 } \right) where t is a real number
    Now for colinearity we must have z_3  = z_1  + s\left( {z_2  - z_1 } \right)\quad  \Rightarrow \quad s = \frac{{z_3  - z_1 }}{{z_2  - z_1 }} for some s.
    But remember that s is real, so {\mathop{\rm Im}\nolimits} \left( {\frac{{z_3  - z_1 }}{{z_2  - z_1 }}} \right) = 0.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Plato View Post
    This may not be different way of looking at the problem.
    The line determined by z_1 \,\& \, z_2 is l(t) = z_1  + t\left( {z_2  - z_1 } \right) where t is a real number
    Now for colinearity we must have z_3  = z_1  + s\left( {z_2  - z_1 } \right)\quad  \Rightarrow \quad s = \frac{{z_3  - z_1 }}{{z_2  - z_1 }} for some s.
    But remember that s is real, so {\mathop{\rm Im}\nolimits} \left( {\frac{{z_3  - z_1 }}{{z_2  - z_1 }}} \right) = 0.
    Ha! Yes, that works for me. Thank you. And the analysis here actually bears some similarity to the next problem in the book.

    -Dan
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