Suppose we are given two points on unit circle which are represented as complex numbers $u$, $v$. We want to show that the intersection of the line through $u$ and $v$ and the real axis is
$$z=\frac{u+v}{1+uv}.$$
(Again, the complex number $z$ represents the point in plane.)
I will show below my derivation of this formula. I would like to ask:
- Are there simpler/other ways to get to the same result?
If I draw the picture, then I see, that I want to find a point which lies on the line going through $a=\frac{u+v}2$ (=center of the line segment joining $u$ and $v$) which is perpendicular to the vector from the origin the point $a$.
If $a$ is the point with coordinates $(R\cos\alpha,R\sin\alpha)$, then I get that $z=\frac{R}{\cos\alpha}$ (simply from the right-angle triangle fromed by $0$, $a$ and $z$). Since $R=|a|$ and $R\cos\alpha=\frac{a+\overline{a}}2$, I get that
$$z=\frac{|a|^2}{\frac{a+\overline{a}}2} = \frac{2a\overline{a}}{a+\overline a}.$$
By plugging $a=\frac{u+v}2$ and using $\overline u = \frac1u$ and $\overline v=\frac 1v$ (since they are on unit circle), I get
$$z = \frac{2a\overline{a}}{a+\overline a} = \frac{\frac{(u+v)(\overline{u}+\overline{v})}2}{\frac{u+\overline{u}+v+\overline{v}}2} =
\frac{(u+v)(\overline{u}+\overline{v})}{u+\overline{u}+v+\overline{v}} = \frac{(u+v)(\frac1u+\frac1v)}{u+\frac1u+v+\frac1v} =
\frac{\frac{(u+v)^2}{uv}}{u+\frac1u+v+\frac1v} = \frac{(u+v)^2}{u^2v+v+uv^2+u} = \frac{(u+v)^2}{(u+v)(uv+1)} = \frac{u+v}{uv+1}.$$
This is the same question as in another post here. The post was put on-hold. Since it seems interesting to me, I would like to repost a better version of that question. Hopefully, I have added enough so that this is not closed for lack of context/effort.
Best Answer
I don't see how complex numbers play a role here. I am just using the first point as $u(\cos\theta_1,\sin\theta_1)$ and the second as $v(\cos\theta_2,\sin\theta_2)$, write down the equation of the line (this assumes that we don't have $\theta_1=\theta_2$ or $\theta_1=-\theta_2$ --- in the first case there is no line and in the second the answer is $\cos(\theta_1)=\cos(\theta_2)$ - draw a picture).
If $\theta_2+\theta_1\neq\pi$ I find the intersection at
$$\frac{\sin(\theta_1-\theta_2)}{\sin(\theta_1)-\sin(\theta_2)}.$$