If all the distinct roots of the equation $x^{47}+2x^{46}+3x^{45}+\cdots \cdots +24x^{24}+23x^{23}+\cdots +2x^2+x=0$ are $z_{1},z_{2},z_{3},\cdots ,z_{k}$ and Imaginary part of $z^2_{k}$ is $b_{k}$. Then value of $|b_{1}|+|b_{2}|+\cdots \cdots +|b_{k}|$ is
Try: Using the face that
$$1+2x+3x^2+\cdots \cdots +nx^{n-1}=\frac{1-x^n}{(1-x)^2}-\frac{nx^n}{1-x}$$
so our expresion is $$x\bigg[x^{46}+2x^{45}+3x^{44}+\cdots \cdots +23 x^{24}\bigg]+\bigg[1+2x+3x^2+\cdots \cdots +24x^{23}\bigg]=0$$
$$x\cdot x^{46}\bigg[1+2x^{-1}+3x^{-2}+\cdots \cdots +23x^{-22}\bigg]+\bigg[1+2x+3x^2+\cdots \cdots +24x^{23}\bigg]=\frac{(1-x^{24})^2}{(1-x)^2\cdot x^{47}}$$
Could some help me how to solve it, Thanks in advanced
If you set $\dfrac{(1-x^{24})^2}{(1-x)^2\cdot x^{47}}$ to $0$ then the solution is just the $24$th roots of unity.
Then $$b_k=\left(\cos\left(\frac{\pi k}{12}\right)+i\sin\left(\frac{\pi k}{12}\right)\right)^2\implies\Im(b_k)=\sin\left(\frac{\pi k}6\right)$$ where $k=1,2,\cdots,23$.
Therefore, $$|b_1|+\cdots+|b_{23}|=\sum_{k=1}^{23}\bigg|\sin\left(\frac{\pi k}6\right)\bigg|$$ which is easy to evaluate since $\sin$ is periodic with period $\pi$.