Going to be beautiful

eqn will open up to 2(Z^7) + 7(Z^6) +. . . =0 (by binomial theorem)

so, Z1 +Z2 +Z3 +Z4 + . . . +Z7 = (-7)/2

hence, summation Im (Zr) = 0 and summation Re(Zr) = (-7)/2

|2a| = 7, |b+9|=9

and from quadratic eqn -c = |2a| + |b+9| =16 and d=|2a|*|b+9|=63 hence |c| + |d| +21 =100

$(z+1)^7=-z^7\\\Rightarrow |z+1|=|z|$

So all of the 7 roots lies on $x=\dfrac{-1}{2}$ . Therefore, $a=\displaystyle\sum_{r=1}^7 Re(z_r)=\dfrac{-7}{2}$ .

By symmetry, $b=\displaystyle\sum_{r=1}^7 Im(z_r)=0$

So, $|2a|=7, |b+9|=9$

Hence the equation is $x^2-16x+63=0$

$16+63+21=100$

A complex numbers solution: (z+1)^7=-z^7. Taking 7th roots, we see that z+1= -wz, where w is a 7th root of unity. We know that there are 7 of these and these are all distinct. Rearranging gives z=-1/(1+w) .

Some experimenting will show us that the real term is always -1/2 so a=-7/2. Now for the imaginary part, we notice that the polynomial has real coefficients so if a+ib is a root, a-ib is also a root. So b=0. Hence, the roots of the quadratic equation are 7 and 9.

(x-7)(x-9)= x^2-16x+63. So the value of the expression required is 16+63+21=100.

This is one of those answers which you immediately know is right even before checking.