Calculation with complex numbers

Question

$4$ complex numbers $a,b,c$ and $d$ have the same, non-zero module $r$. Find the result of the following expression: $$|\frac{abc+abd+acd+bcd}{a+b+c+d}|$$ I think this task is supposed to be solved by using trigonometrical expressions of complex numbers but that led me nowhere.

Answer 1

Consider the case $r=1$. Let $z=abcd$. Then the expression you have is $$\left|z\cdot\frac{1/a+1/b+1/c+1/d}{a+b+c+d}\right|=\left|z\right|\cdot\left|\frac{\bar a+\bar b+\bar c+\bar d}{a+b+c+d}\right|=\left|\frac{\bar a+\bar b+\bar c+\bar d}{a+b+c+d}\right| $$ where $\bar k$ dentoes the complex conjugate of $k$. Note that this is true because the complex conjugate of a complex number with modulus $1$ is exactly its reciprocal. But for any complex numbers $z_1,z_2$ we have $\overline{z_1+z_2}=\bar z_1+\bar z_2$, so this is actually just $$\left|\frac{\overline{a+b+c+d}}{a+b+c+d}\right|=\frac{|\overline{a+b+c+d}|}{|a+b+c+d|}=1.$$ Can you extend this approach to $r\neq 1$?