I am faced with the following problem..
Consider three points $A (a), B (b), C(c)$ on the unit circle $|a|= |b|= |c|=1$. Find the complex coordinates of the point $D (d)$, where $D$ also lies on the unit circle & $AD \perp BC$.
Here's what I did:
By the first condition, $|d|=1$. We can think this like a cartesian coordinate system where the coordinates of $z$ are $ (\Re (x), \Im (x))$. So the perpendicular condition gives us $\frac{\Im (d)- \Im (a)}{\Re (d) - \Re (a)} \frac{\Im (b) - \Im (c)}{\Re (b) - \Re (c)}= -1$. But how do i continue from here? I am trying to find $d$ explicitly in terms of $a, b, c$. Expanding this looks huge.
Thanks.
a point $D$ on $BC$ can be written as $d = (1-t)b + tc$ for some real number $t.$ now we will use the fact that $a$ and $b$ are orthogonal iff $a\bar b + \bar a b = 0.$ we need $t$ so that $AD$ is orthogonal to $BC,$ so
$$[(1-t)b + tc - a][\bar b-\bar c] + [(1 -t)\bar b + t \bar c - \bar a](b- c) = 0 $$
can you take it from here?
if you want the point $D$ on the circumference such that $AD$ and $BC$ are orthogonal then just use $$(a-d)(\bar b - \bar c) + (\bar a - \bar d)(b -c) = 0$$ and the fact that $\bar d = \frac{1}{d}$ to get a quadratic equation for $d.$
solving for $d$ you find $d = a, d = \dfrac{b - c}{a(\bar b - \bar c)}.$ the second is the you want.
check: suppose $a = i, b = -1, c = 1,$ then we know $d = -i.$ let us check the formula for $d = -2/(-2i) = -i$