# Thread:

1. ## [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

2. Originally Posted by topsquark
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

3. Originally Posted by CaptainBlack
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

4. 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 ${\mathop{\rm Im}\nolimits} \left( {\frac{{z_3 - z_1 }}{{z_2 - z_1 }}} \right) = 0$.

5. Originally Posted by Plato
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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