Let $z_k(k = 0, 1, 2, ............ 6)$ be the roots of the equation $(z + 1)^7 + z^7 = 0$, then $\sum { k=0 }^{ 6 }{ Re({ z }{ k }) } $ is equal to
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Let $z_k(k = 0, 1, 2, ............ 6)$ be the roots of the equation $(z + 1)^7 + z^7 = 0$, then $\sum { k=0 }^{ 6 }{ Re({ z }{ k }) } $ is equal to
The equation can be rewritten as ((z + 1)/z)^7 = -1. This implies that the modulus of (z + 1)/z is 1, which geometrically means that z is equidistant from 0 and -1, so its real part must be -1/2. Since the equation is of degree 7, there are 7 roots, and the sum of their real parts is 7 * (-1/2) = -7/2.
By expanding the given equation (z + 1)^7 + z^7 = 0, we get the polynomial 2z^7 + 7z^6 + 21z^5 + 35z^4 + 35z^3 + 21z^2 + 7z + 1 = 0. We are tasked with finding the sum of the real parts of its seven roots, which is equivalent to finding the real part of the sum of the roots. Using Vieta's formulas on the expanded polynomial, the sum of the roots z0 through z6 is given by the negative coefficient of the z^6 term divided by the leading coefficient. Therefore, the sum of the roots is -7 / 2, and since this value is purely real, the sum of the real parts is -7 / 2.