Let $$\omega$$ be a nonreal root of $$z^3 = 1.$$ Let $$a_1,$$ $$a_2,$$ $$\dots,$$ $$a_n$$ be real numbers such that $$\frac{1}{a_1 + \omega} + \frac{1}{a_2 + \omega} + \dots + \frac{1}{a_n + \omega} = 2 + 5i.$$Compute $$\frac{2a_1 - 1}{a_1^2 - a_1 + 1} + \frac{2a_2 - 1}{a_2^2 - a_2 + 1} + \dots + \frac{2a_n - 1}{a_n^2 - a_n + 1}.$$

I have no clue how to do this. Can someone help?

• If you trust the problem setter to have made sure there is a unique solution, you can choose $n=1$ and your favorite root for $\omega$. Then is is routine algebra to solve for $a_1$ and plug into the second expression. – Ross Millikan Apr 23 at 21:36
• @RossMillikan Unfortunately that $a_1$ is not real. – peterwhy Apr 23 at 23:45
• @peterwhy: why is that an issue? We are just solving a linear equation, then doing complex arithmetic. But your solution is clearly the intended one. – Ross Millikan Apr 23 at 23:49

Hint: $$\overline \omega$$ is another non-real root of $$z^3=1$$,

$$\sum_{j=1}^n\frac1{a_j+\overline\omega} = \overline{2+5i}$$

$$1,\omega, \overline\omega$$ are roots of $$z^3-1 = 0$$, so

$$\omega\overline\omega = 1,\quad \omega+\overline\omega + 1 = 0$$

Consider the following sum,

\begin{align*} \sum_{j=1}^n\frac1{a_j+\omega} + \sum_{j=1}^n\frac1{a_j+\overline\omega} &= \sum_{j=1}^n\frac{(a_j+\omega)+(a_j+\overline\omega)}{(a_j+\omega)(a_j+\overline\omega)}\\ &= \sum_{j=1}^n\frac{2a_j+\omega+\overline\omega}{a_j^2 + \omega a_j +\overline\omega a_j + \omega\overline\omega}\\ &= \sum_{j=1}^n\frac{2a_j-1}{a_j^2 -a_j + 1}\\ \end{align*}

Rationalize the denominator for each term of the left hand side & the right hand side

For integer $$n$$ not divisible by$$3$$

Let $$p=\dfrac{2n\pi}3,2\cos p=-1\ \ \ \ (1)$$

$$\dfrac1{a_j+w}=\dfrac1{a_j+\cos p+i\sin p}=\dfrac{(a_j+\cos p)-i\sin p}{a_j^2+2a_j\cos p+1}$$

Use $$(1)$$

Equate the real parts