3

I've encountered this problem recently: Let $z_1,z_2,z_3\in\mathbb C$ be three distinct points and L be the line segment joining $z_1,z_3$. Let the straight line passing through $z_2$ meet the line L perpendicularly at w. Find w in terms of $z_1,z_2,z_3$.

My attempt: Let's translate the whole thing so that $z_3$ is the origin $0$. L is the line segment joining this point and $z_1$. $z_2$ is still there. Then w=k$z_1$ for some k $\in\mathbb R$.

$$\frac{z_2-w}{w}$$ is purely imaginary. so $$-\frac{z_2-w}{w}=\frac{\bar z_2-\bar w}{\bar w}$$ $$-\frac{z_2-kz_1}{kz_1}=\frac{\bar z_2-k\bar z_1}{k\bar z_1}$$ ... $$2kz_1 \bar z_1=z_1 \bar z_2+\bar z_1z_2$$ $$2k|z_1|^2=2Re(z_1 \bar z_2)$$ $$k=\frac{Re(z_1 \bar z_2)}{|z_1|^2}$$ Therefore $$w=\frac{Re(z_1 \bar z_2)}{|z_1|^2}z_1$$ This makes total sense to me. What if $z_3$ is not translated to the origin?

1 Answers1

0

If $L'$ is the line segment joining $z_1-z_3$ and $0$, then the line passing through $z_2-z_3$ meets $L'$ at $w-z_3$. Your work implies that $$w=\frac{Re((z_1-z_3)(\bar z_2-\bar z_3))}{|z_1-z_3|^2}(z_1-z_3)+z_3$$

CY Aries
  • 23,393