Geometry with complex numbers. Let $a$, $b$, $c$, and $d$ be four complex numbers on the unit circle, such that the line joining $a$ and $b$ is perpendicular to the line joining $c$ and $d$.

Find a simple expression for $d$ in terms of $a$, $b$, and $c$.
I have thought of using power of a point, but that has gotten me nowhere so far.  Can anyone help me? 
 A: We want that 
$$\lambda:={d-c\over b-a}$$
is purely imaginary. This is equivalent with $\lambda=-\bar\lambda$, or
$${d-c\over b-a}=-{\bar d-\bar c\over\bar b-\bar a }=-{{1\over d}-{1\over c}\over{1\over b}-{1\over a}}\ .$$
This at once simplifies to $ab=-cd$.
A: The angles corresponding to the arcs from $d$ to $a$ and from $c$ to $b$ add up to $180$ degrees.
That is,
$$\frac{ab}{dc}=-1$$
A: Since $|a| = 1$, we have $a \overline{a} = |a|^2 = 1$, so $\overline{a} = 1/a$. Similarly, $\overline{b} = 1/b$, $\overline{c} = 1/c$, and $\overline{d} = 1/d$.
The line joining $a$ and $b$ is perpendicular to the line joining $c$ and $d$, so
$(b - a)(\overline{d} - \overline{c}) + (\overline{b} - \overline{a})(d - c) = 0,$
which becomes
\begin{align*}
(b - a) \left( \frac{1}{d} - \frac{1}{c} \right) + \left( \frac{1}{b} - \frac{1}{a} \right) (d - c) &= 0 \\
\Rightarrow \quad \frac{(b - a)(c - d)}{cd} + \frac{(a - b)(d - c)}{ab} &= 0 \\
\Rightarrow \quad (b - a)(c - d) \left( \frac{1}{cd} + \frac{1}{ab} \right) &= 0.
\end{align*}
Since $b \neq a$ and $c \neq d$, we may divide both sides by $(b - a)(c - d)$, to get
$\frac{1}{cd} + \frac{1}{ab} = 0.$
Solving for $d$, we find $d = \boxed{-ab/c}$.
