If two of the complex numbers (a-b)/(c-d) , (b-c)/(a-d)…

If two of the complex numbers $\dfrac{a-b}{c-d}$, $\dfrac{b-c}{a-d}$, $\dfrac{c-a}{b-d}$ are purely imaginary, show that the third is also purely imaginary.

Step 1 · Name the three ratios

$$ u=\frac{a-b}{c-d}, \qquad v=\frac{b-c}{a-d}, \qquad w=\frac{c-a}{b-d} $$

Step 2 · The key identity

Clearing denominators and expanding (using that the numerators $a-b,\,b-c,\,c-a$ sum to zero) gives the identity:

$$ \boxed{u+v+w+uvw=0} $$

valid for all complex $a,b,c,d$ with the denominators nonzero. (This can be checked directly: multiplying out $u+v+w+uvw$ over the common denominator $(c-d)(a-d)(b-d)$, every term cancels.)

Step 3 · Suppose $u,v$ are purely imaginary

By definition, a purely imaginary number equals the negative of its own conjugate: $\bar u=-u$, $\bar v=-v$.

Step 4 · Conjugate the identity

$$ \bar u+\bar v+\bar w+\bar u\bar v\bar w=0 $$

Substituting $\bar u=-u$, $\bar v=-v$:

$$ -u-v+\bar w+(-u)(-v)\bar w=0 \quad\Longrightarrow\quad -u-v+\bar w(1+uv)=0 \quad\Longrightarrow\quad \bar w=\frac{u+v}{1+uv} $$

(assuming $1+uv\neq0$; see Step 6).

Step 5 · Solve the original identity for $w$

From $u+v+w+uvw=0$: $\ w(1+uv)=-(u+v)$, so:

$$ w=\frac{-(u+v)}{1+uv} $$

Comparing with Step 4:

$$ \bar w=\frac{u+v}{1+uv}=-\,\frac{-(u+v)}{1+uv}=-w $$

$$ \boxed{\bar w=-w} $$

which says exactly that $w$ is purely imaginary.

Step 6 · The excluded case $1+uv=0$

If $u=i\alpha$, $v=i\beta$ with $\alpha,\beta$ real, then $uv=-\alpha\beta$ is real, and $1+uv=0$ would require $\alpha\beta=1$. Combined with $w(1+uv)=-(u+v)$, this forces $u+v=0$, i.e. $\alpha=-\beta$; together with $\alpha\beta=1$ this gives $-\alpha^2=1$, impossible for real $\alpha$. So this case cannot actually occur when $u,v$ are purely imaginary, and the argument in Steps 4–5 always applies.

Step 7 · By symmetry

The identity $u+v+w+uvw=0$ is symmetric under simultaneously cycling $(a,b,c)\to(b,c,a)$, which cycles $(u,v,w)\to(v,w,u)$. So the same argument, applied with any two of $u,v,w$ assumed purely imaginary, shows the third is too — the roles of $u,v,w$ in Steps 3–5 can be permuted freely.