জ্যামিতিতে জটিল সংখ্যা: একক বৃত্তের জ্যায়ের উপর প্রক্ষেপণ

Suppose we have a unit circle. Points P,Q are on the circle as well as a complex number w. We want to find the point K on PQ such PQ is perpendicular to KW

Yes, it's better to find the equation first, and then use complex numbers.

Let $P=a+bi,Q=c+di,w=e+fi, |p|=|q|=|w|=1$

First we will express everything in terms of p,q. Later we will use a,b,c,d,e,f.

Each point on PQ has the form $mP+(1-m)Q$ where m is between 0 and 1. Let the point K we want have value $m=k$. Notice the capital/lowercase difference.

We will not do the substitution until it is absolutely necessary, because the substitution is messy.

So... if $PQ$ is perpendicular to $KW$, then $\overline{\frac{p-q}{w-K}} = \frac{q-p}{w-K}$.

The conjugate takes the conjugate of the entire fraction. Because of the conjugate identities (ex: $\overline{a+b}=\overline{a}+\overline{b}$ we can rewrite this as

$\frac{\overline{p}-\overline{q}}{\overline{w}-\overline{K}} = \frac{q-p}{w-K}$.

Here comes the clever part...

The reason is that $x \overline{x} = |x|^2 = 1$.

Now, we can rewrite the fraction on the left because q,p,w are all on the unit circle. We get..

$pq(1 - w \overline{K}) = w^2 - wK$.

Now $K = kp + (1-k)q$, so we can plug that in.

$pq(1- w \overline{kp+(1-k)q}) = w^2 - wkp + (1-k)wq$.

How do we find the conjugate of K? We'll need to use $p=a+bi, q= c+di$. For simplicity's sake, I will **only plug these in for the conjugate K term. Everywhere else, the p and q will stay as they are **.

So then $pq(1-w \overline{k (a+bi) + (1-k)(c+di)}) = w^2 - wkp + wq - kwq$

We can expand and solve for k to get

$k = -\frac{pq-pqwi+pqwdi-w^2-wa}{wp+wq+ipqwci+bi-di-a}$

Not exactly the nicest thing ever... I prefer to ignore this and just use

$\boxed{pq(1 - w \overline{K})= w^2 - wK}$ when solving problems, as it is much simpler.

Note: I apologize if I made an arithmetic error or two along the way. Given the messiness of this formula, it is almost inevitable