Show that $(a-b)(c-d)(\bar{a}-\bar{d})(\bar{c}-\bar{b})+i(c\bar{c}-d\bar{d})\text{Im}(c\bar{b}-c\bar{a}-a\bar{b})$ is real This is an exercise in Remmert's Theory of Complex Functions, GTM 122, page 17. 

Show that for $a,b,c,d\in\mathbb{C}$ with $\lvert a\rvert=\lvert b\rvert=\lvert c\rvert$ the complex number $$(a-b)(c-d)(\bar{a}-\bar{d})(\bar{c}-\bar{b})+i(c\bar{c}-d\bar{d})\text{Im}(c\bar{b}-c\bar{a}-a\bar{b})$$ is real.

I let the above expression $w$ and tried showing the $w-\bar{w}=0$, but the expression is too complicated to handle and it seems like $w-\bar{w}$ depends on $d$, also. Is there any simpler way of showing $w$ is real? Please enlighten me.
 A: If $b=0$ or $c=0$, then the complex number equals $0$ which is real.
In the following, $b\not =0$ and $c\not=0$.
In order to prove that the claim is true, it is sufficient to prove the following two lemmas :
Lemma 1 : $$\text{Im}(c\bar{b}-c\bar{a}-a\bar{b})=-\frac{\bar a(a-b)(b-c)(c-a)}{2bc}$$
Lemma 2 : $$\text{Im}((a-b)(c-d)(\bar{a}-\bar{d})(\bar{c}-\bar{b}))=(c\bar c-d\bar d)\times \frac{\bar a(a-b)(b-c)(c-a)}{2bc}$$
From the two lemmas, we see that the complex number equals $\text{Re}((a-b)(c-d)(\bar{a}-\bar{d})(\bar{c}-\bar{b}))$ which is real, so the claim is true.

Proof for lemma 1 : 
Using $a\bar a=b\bar b=c\bar c$, we have 
$$\begin{align}\text{Im}(c\bar{b}-c\bar{a}-a\bar{b})
&=\frac 12\left((c\bar{b}-c\bar{a}-a\bar{b})-\overline{(c\bar{b}-c\bar{a}-a\bar{b})}\right)
\\\\&=\frac 12\left((c\bar{b}-c\bar{a}-a\bar{b})-(\bar c b-\bar ca-\bar a b)\right)
\\\\&=\frac 12\left(c\frac{a\bar a}{b}-c\bar{a}-a\frac{a\bar a}{b}-\frac{a\bar a}{c} b+\frac{a\bar a}{c}a+\bar a b\right)
\\\\&=\frac{\bar a}{2bc}\left(c^2a-bc^2-ca^2-ab^2+a^2b+b^2c\right)
\\\\&=\frac{\bar a}{2bc}\left(c^2(a-b)-c(a-b)(a+b)+ab(a-b)\right)
\\\\&=\frac{\bar a(a-b)}{2bc}\left(c^2-c(a+b)+ab\right)
\\\\&=\frac{\bar a(a-b)}{2bc}(c-a)(c-b)
\\\\&=-\frac{\bar a(a-b)(b-c)(c-a)}{2bc}\qquad\square\end{align}$$
Proof for lemma 2 : 
Using $a\bar a=b\bar b=c\bar c$, we have 
$$\begin{align}&\text{Im}((a-b)(c-d)(\bar{a}-\bar{d})(\bar{c}-\bar{b}))
\\\\&=\frac 12(a-b)(c-d)(\bar{a}-\bar{d})(\bar{c}-\bar b)-\frac 12\overline{(a-b)(c-d)(\bar{a}-\bar{d})(\bar{c}-\bar{b})}
\\\\&=\frac 12(a-b)(c-d)(\bar{a}-\bar{d})(\bar{c}-\bar{b})-\frac 12(\bar a-\bar b)(\bar c-\bar d)(a-d)(c-b)
\\\\&=\frac 12(a-b)(c-d)(\bar{a}-\bar{d})\left(\frac{a\bar a}{c}-\frac{a\bar a}{b}\right)-\frac 12\left(\bar a-\frac{a\bar a}{b}\right)\left(\frac{a\bar a}{c}-\bar d\right)(a-d)(c-b)
\\\\&=\frac{a\bar a}{2bc}(a-b)(c-d)(\bar{a}-\bar{d})\left(b-c\right)-\frac{\bar a}{2bc}\left(b-a\right)\left(a\bar a-c\bar d\right)(a-d)(c-b)
\\\\&=\frac{\bar a(a-b)(b-c)}{2bc}\left(a(c-d)(\bar{a}-\bar{d})-c(\bar c-\bar d)(a-d)\right)
\\\\&=\frac{\bar a(a-b)(b-c)}{2bc}\left((c-a)c\bar c-(c-a)d\bar d\right)
\\\\&=\frac{\bar a(a-b)(b-c)(c-a)(c\bar c-d\bar d)}{2bc}
\\\\&=(c\bar c-d\bar d)\times \frac{\bar a(a-b)(b-c)(c-a)}{2bc}\qquad\square\end{align}$$
