If $a_{k} = {cosα}_{k} + i {sinα}_{k}, : k = 1, 2, 3$ and $a_{1}, a_{2}, a_{3}$ are the roots of the equation $x^{3} + bx + c = 0,$ then the real part of b =

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$\frac{2}{3}$

-1

answer is C.

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Detailed Solution

$a_{k} = cis α_{k}$

$a_{1}, a_{2}, a_{3}$ are roots of $x^{3} + bx + c = 0$

$\begin{array}{rcl} ∴ a_{1} + a_{2} + a_{3} & = & 0, a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} = b \end{array}$

$\begin{array}{rcl} a_{1} + a_{2} + a_{3} & = & 0 \Rightarrow cis α_{1} + cis α_{2} + cis α_{3} = 0 \end{array}$

$\begin{array}{rcl} \Rightarrow & (\cos α_{1} + \cos α_{2} + \cos α_{3}) + i (\sin α_{1} + \sin α_{2} + \sin α_{3}) = 0 \\ \Rightarrow & (\cos α_{1} + \cos α_{2} + \cos α_{3}) - i (\sin α_{1} + \sin α_{2} + \sin α_{3}) = 0 \\ ∴ \frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} & = & 0 \\ a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} & = & 0 \\ \Rightarrow & b = 0 \\ Re (b) & = & 0 \end{array}$