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How to show that two complex numbers lie on different half planes determined by a line?

In particular, I need to show that $z$ and $w$ lie on different half planes determined by the line passing through $z_1$ and $z_2.$

I need some formula in terms of $z, w, z_1, z_2$. Any hint will be appreciated.Thanks

User
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  • Are all numbers given to you on the form $x + iy$? – Arthur Sep 08 '17 at 12:39
  • (+1) because I often wondered how to determine on which "side" of a line a specific point is. – Peter Sep 08 '17 at 12:42
  • Are you familiar with $ℝ^2$? Because $ℂ$ can easily be seen as $ℝ^2$. So, given a point and a straight line in the plane $ℝ^2$, do you know how to determine if the point is on one or the other side? If no, have a look at this question – P. Siehr Sep 08 '17 at 12:43
  • Nothing complex in here (with both meanings for "complex") : take the perpendicular bissector of the two points, and it's done ! – Jean Marie Sep 08 '17 at 12:48
  • Great, then for a given number $z∈ℂ$ with $z=z_1+z_2i$ you can just use a vector $\mathbb{z}:=\begin{pmatrix}z_1 \ z_2 \end{pmatrix}∈ℝ^2$ or a point $(z_1,z_2)$ [whatever you are more familiar with]. Now your problem is the same, but written in "the terms of $ℝ^2$". – P. Siehr Sep 08 '17 at 12:51
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    given any two distinct points $z_1, z_2 \in \mathbb{C}$, a third point $z \in \mathbb{C}$ lies on the line joining $z_1, z_2$ iff $\Re \frac{z-z_2}{z_1 - z_2} = 0$. two points $z, w$ falling on the different sides of the line iff $\Im \frac{z-z_2}{z_1 - z_2}$ and $\Im\frac{w-z_2}{z_1-z_2}$ are both non-zero but have different signs. The subtraction of $z_2$ from both denominator and numerator corresponds to moving the origin to $z_2$. The division by $z_1 - z_2$ corresponds to rotating and scaling the complex plane so that $z_1 - z_2$ becomes the unit vector in $x$-direction. – achille hui Sep 08 '17 at 13:10
  • @achillehui, How do we get $\Re \frac{z-z_2}{z_1-z_2}=0$ – User Sep 08 '17 at 13:13
  • @achillehui, Can you suggest me any book or paper, from where I can read them? – User Sep 08 '17 at 13:16
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    Look at first chapter of Roland Deaux's book "Introduction to the geometry of complex numbers". It has some examples about what common arithmetic operations in complex numbers means in geometry. – achille hui Sep 08 '17 at 13:23
  • @achillehui, Thanks for suggesting such a good book. – User Sep 08 '17 at 15:35

3 Answers3

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$\omega$ and $\xi$ are on opposite sides with respect to the line $\ell_{z_1,z_2}$ through $z_1$ and $z_2$ if and only if $$\Im((\omega-z_1) (\overline{z_2}-\overline{z_1}))\cdot\Im((\xi-z_1) (\overline{z_2}-\overline{z_1}))<0\tag{1}$$ Indeed if $$f:\mathbb{C}\to\mathbb{R},f(t)=\Im((t-z_1) (\overline{z_2}-\overline{z_1}))$$ then this is a continuous function, it assumes the value $0$ if and only if $(t-z_1) (\overline{z_2}-\overline{z_1})$ is real, or equivalently $\dfrac{t-z_1}{z_2-z_1}\in\mathbb{R}$ or $t,z_1,z_2$ are on the same line, that is $t\in\ell_{z_1,z_2}$.

It follows that $f$ keeps a constant signs on each of the half spaces determined by $\ell_{z_1,z_2}$. Testing $f(z_1\pm i(z_2-z_1))$ we see that these signs are opposite. So, condition $(1)$ is just saying that $f(\omega)f(\xi)<0$ so $\omega$ and $\xi$ are on opposite sides with respect to the line $\ell_{z_1,z_2}$.

Felix Klein
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This is perhaps done best in the Euclidean plane rather than complex plane (not that there's much of an actual difference, but how we think about it and write stuff is different).

First, find an equation for your line, and write it on the form $ax + by + c = 0$. Then the two halves of the plane are distinguished by what sign you get if you evaluate the left-hand side of that equation. For instance, say $z = (x_1, y_1)$ and $w = (x_2, y_2)$. Then $z$ and $w$ lie on the same side of the line iff $ax_1+by_1 + c$ has the same sign as $ax_2 + by_2 + c$.

Arthur
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(Notation note:$\;d·e\quad$denotes the e-coördinate of point d.)

If $\;\begin{vmatrix} z_1.x&z_1.y&1\\z_2.x&z_2.y&1\\z.x&z.y&1 \end{vmatrix}\quad$ and $\;\begin{vmatrix} z_1.x&z_1.y&1\\z_2.x&z_2.y&1\\w.x&w.y&1 \end{vmatrix}\quad$ have differing signs,

then the points bracket the line.

Donn Miller
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