Unit Circle and Imaginary Axis geometry problem

Question

Suppose $x$ and $y$ are points on the unit circle such that the line through $x$ and $y$ intersects the real axis. Show that if $z$ is the point where this line intersects the real axis, then $z = \dfrac{x+y}{xy+1}$.

How should I go about attempting this problem? I am currently struggling to find a starting point (and going through with the entire proof).

Any help is appreciated!

Answer 1

The points $x$, $y$ and $z$ are col-linear iff $$\operatorname{Im}(\overline{x}y+\overline{y}z+\overline{z}x) = 0$$ $$\overline{x}y+\overline{y}z+\overline{z}x-\overline{y}x-\overline{z}y-\overline{x}z = 0$$ since $z=\bar{z}$, $\bar{x}=\dfrac{1}{x}$ and $\bar{y}=\dfrac{1}{y}$ \begin{eqnarray} z(\overline{y}+x-y-\overline{x})&=& -\overline{x}y+\overline{y}x\\ z(x-y+\dfrac{x-y}{xy}) &=& \dfrac{x}{y}-\dfrac{y}{x}\\ z(\dfrac{(x-y)(1+xy)}{xy})&=& \dfrac{(x-y)(x+y)}{xy}\\ z &=& \dfrac{x+y}{1+xy} \end{eqnarray}

Answer 2

It suffices to show that $\frac{x+y}{xy+1}$ is both real and on the line connecting $x$ and $y$. For the first one, check that $\frac{x+y}{xy+1}$ minus its complex conjugate equals $0$ (hint: if $x,y$ are on the unit circle, what's another way to write $\bar x,\bar y$?). For the second one, simply compare slopes between pairs of points (which is essentially checking the cross ratio, with the fourth point at $\infty$).

Answer 3

Hints: $z$ is on the line through $\;x,y\;$ iff $\;\exists \lambda \in \mathbb{R}\,$ so that $\,z = \lambda x + (1-\lambda)y\,$.

• Determine $\lambda$ from the condition that $z$ lies on the real axis, which is equivalent to $z=\bar z\,$:

$$ \lambda x + (1-\lambda)y = \lambda \bar x + (1-\lambda) \bar y \quad\iff\quad \lambda = \cfrac{-(y - \bar y)}{x - \bar x-(y - \bar y)} $$

• Note that $1-\lambda=\cfrac{x - \bar x}{x - \bar x-(y - \bar y)}$, and calculate $z\,$:

$$ \require{cancel} \begin{align} z = \lambda x + (1-\lambda)y = \cfrac{-x(\cancel{y} - \bar y) + y(\cancel{x} - \bar x)}{x - \bar x-(y - \bar y)} = \cfrac{x \bar y - \bar x y}{x - \bar x-(y - \bar y)} \end{align} $$

• Use the condition that $x,y$ are on the unit circle, so substitute $\bar x = 1/x$ and $\bar y = 1/y\,$ in the expression above, then collect and simply.

Answer 4

Notice from the diagram that:

$$ |z|^2 = \left |\frac{x+y}{2}\right |^2 + \left |\frac{x+y}{2} - z\right |^2 \\ \implies zz^* = \frac14 (x+y)(x^* + y^*) + \left[\frac{x+y}{2}-z\right]\left[\frac{x^* + y^*}{2} - z^*\right] $$

Using $x^*=\frac1x$, $y^*=\frac1y$ and $z = z^*$, we get,

$$ z\left(x + y + \frac1x + \frac1y\right) = (x+y)\left(\frac1x + \frac1y\right) $$

which simplifies to

$$ z = \frac{x+y}{xy+1}. $$