Just a warmup

We can easily get the answer from the fact that sum of the $n^{th}$ root of unity is always equals to 0 , when $n>1$ .

As the real root is 4 , sum of the complex roots will be, $\boldsymbol{0-4=}\color{#69047E}{\boxed{-4}}$

x^5=1024 x^5-1024=0 (x-4)(x^4+4x^3+16x^2+64x+256)=0 as is observed the only real root of the equation is 4 hence all the complex roots are solutions of x^4+4x^3+16x^2+64x+256=0 further we know that the sum of roots of an equation of any degree is -b/a hence sum of complex roots is-4

Let $z=\sqrt[5]{1024},$ implying $z$ is a $5^{th}$ root of unity

By the De Moivres Theorem, we have $z^n=(r(e^{i\theta}))^n=r^n(e^{\frac{2k\pi}ni})$ for $k=0,1,2,\ldots,n-1$

The $5^{th}$ roots of unity are $4e^{\frac{2k\pi}5i}$ for $k=0,1,2,3,4.$

This gives the roots of unity $4,4e^{\frac{2\pi}5i},4e^{\frac{4\pi}5i},4e^{\frac{6\pi}5i},4e^{\frac{8\pi}5i}$

Thus, the sum of all complex roots is

$\begin{aligned}4e^{\frac{2\pi}5i}+4e^{\frac{4\pi}5i}+4e^{\frac{6\pi}5i}+4e^{\frac{8\pi}5i}=&\ \dfrac{4e^{\frac{2\pi}5i}\left[\left(e^{\frac{2\pi}5i}\right)^4-1\right]}{e^{\frac{2\pi}5i}-1}\\=&\ \dfrac{4e^{\frac{2\pi}5i}(e^{\frac{8\pi}5i})-4e^{\frac{2\pi}5i}}{e^{\frac{2\pi}5i}-1}\\=&\ \dfrac{4e^{2\pi i}-4e^{\frac{2\pi}5i}}{e^{\frac{2\pi}5i}-1}\\=&\ \dfrac{4(1-e^{\frac{2\pi}5i})}{e^{\frac{2\pi}5i}-1}\\=&\ \dfrac{-4(e^{\frac{2\pi}5i}-1)}{e^{\frac{2\pi}5i}-1}\\=&\ \boxed{-4}\end{aligned}$