Complex Numbers: Show that a given expression is real Let $a,b,c$ be complex numbers such that $|a|=|b|=|c|=1$. Define the complex numbers $m$ and $n$ by $m=a+r(b-a)$ and $n=a+s(c-a)$ where $r$ and $s$ are non zero real numbers. Given that $m/n$ is a real number, prove that $$\frac{(b-a)(b+n)}{(c-a)(c+m)}$$
is also real.
Work: I tried plugging the values of $m$ and $n$ which gives 
$$\frac{(b-a)(b+n)}{(c-a)(c+m)}=\frac{b^2-a^2+s(bc-ab-ca+a^2)}{c^2-a^2+r(bc-ab-ca+a^2)}.$$
However i'm not really sure how to proceed from here. Any hints or solution is appreciated. 
 A: HINT.- After quite tedious calculations I realized that a suitable way is using exponentials. I give here only  the half of the calculation because consider I payed enough attention to this problem.
$$a=\cos\alpha+i\sin\alpha \\b=\cos \beta+i\sin\beta\\c=\cos\gamma+i\sin\gamma$$
$$b-a=(\cos\beta-\cos\alpha)+i(\sin\beta-\sin\alpha)$$ $$b-a=-2\sin\frac{\alpha+\beta}{2}+2\sin\frac{\alpha-\beta}{2}+i\left(2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}\right)$$
$$b-a=2i\sin\frac{\alpha-\beta}{2}\left(\cos\frac{\alpha+\beta}{2}+i\sin\frac{\alpha+\beta}{2}\right)=2i\sin\frac{\alpha-\beta}{2}\space
 \large e^{\frac{i(\alpha+\beta)}{2}}$$
Similarly $$c-a=2i\sin\frac{\alpha-\gamma}{2}\space
 \large e^{\frac{i(\alpha+\gamma)}{2}}$$ Hence
$$\frac{b-a}{c-a}=\frac{\sin\frac{\alpha-\beta}{2}}{\sin\frac{\alpha-\gamma}{2}}\large e^{\frac{i(\beta-\gamma)}{2}}$$
What follows is similar.
A: The statement doesn't hold true.
Counterexample: $a=1, b=-1, c=i, r = s = \frac{1}{2} \implies m = 0, n = \frac{1+i}{2}$ with $\frac{m}{n} = 0 \in \mathbb{R}$. Then:
$$
\frac{(b-a)(b+n)}{(c-a)(c+m)} = \frac{(-1-1)(-1 + \frac{1+i}{2})}{(i-1)\cdot i} = \frac{ -2 \frac{i-1}{2}}{(i-1) \cdot i} = \frac{-1}{i}=i \;\;\not \in\; \mathbb{R}
$$
