If a_k = cos alpha_k + i sin alpha_k for k = | vedclass

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(A) Given that $a_1, a_2, a_3$ are roots of $x^3 + bx + c = 0$. By Vieta's formulas,the sum of roots is $a_1 + a_2 + a_3 = 0$. Since $a_k = \cos \alph

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(A) Given that $a_1, a_2, a_3$ are roots of $x^3 + bx + c = 0$. By Vieta's formulas,the sum of roots is $a_1 + a_2 + a_3 = 0$. Since $a_k = \cos \alpha_k + i \sin \alpha_k$,we have: $(\cos \alpha_1 + \cos \alpha_2 + \cos \alpha_3) + i(\sin \alpha_1 + \sin \alpha_2 + \sin \alpha_3) = 0$. This implies $\sum \cos \alpha_k = 0$ and $\sum \sin \alpha_k = 0$. Also,the coefficient of $x^2$ is $0$,so $\sum a_i = 0$. The coefficient $b$ is the sum of the products of roots taken two at a time: $b = a_1 a_2 + a_2 a_3 + a_3 a_1$. Since $a_k = e^{i \alpha_k}$,we have $b = e^{i(\alpha_1 + \alpha_2)} + e^{i(\alpha_2 + \alpha_3)} + e^{i(\alpha_3 + \alpha_1)}$. Using the property that if $\sum e^{i \alpha_k} = 0$,then $\sum e^{-i \alpha_k} = 0$,we note that $b = \sum a_1 a_2 = \sum \frac{a_1 a_2 a_3}{a_3} = \sum \frac{-c}{a_3} = -c \sum \frac{1}{a_3} = -c \sum \bar{a_3} = -c(0) = 0$. Wait,let's re-evaluate: $b = a_1 a_2 + a_2 a_3 + a_3 a_1$. Since $a_1+a_2+a_3=0$,$(a_1+a_2+a_3)^2 = a_1^2+a_2^2+a_3^2 + 2(a_1 a_2 + a_2 a_3 + a_3 a_1) = 0$. Thus $b = -\frac{1}{2}(a_1^2 + a_2^2 + a_3^2)$. Since $|a_k| = 1$,$a_k^2 = e^{i 2 \alpha_k} = \cos 2 \alpha_k + i \sin 2 \alpha_k$. Given $\sum \cos \alpha_k = 0$ and $\sum \sin \alpha_k = 0$,it is known that for such complex numbers on the unit circle,$\sum a_k^2 = 0$. Therefore,$b = 0$.

If $a_k = \cos \alpha_k + i \sin \alpha_k$ for $k = 1, 2, 3$ and $a_1, a_2, a_3$ are the roots of the equation $x^3 + bx + c = 0$,then the real part of $b$ is:

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