If z is a non-real root of x^7=1,then 1+3z+5z | vedclass

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(B) Let $S = 1+3z+5z^2+7z^3+9z^4+11z^5+13z^6$. Multiplying by $z$,we get $zS = z+3z^2+5z^3+7z^4+9z^5+11z^6+13z^7$. Since $z$ is a root of $x^7=1$,we h

Vedclass · Accepted Answer

$\frac{-14}{1-z}$

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(B) Let $S = 1+3z+5z^2+7z^3+9z^4+11z^5+13z^6$. Multiplying by $z$,we get $zS = z+3z^2+5z^3+7z^4+9z^5+11z^6+13z^7$. Since $z$ is a root of $x^7=1$,we have $z^7=1$. Subtracting the two equations: $(1-z)S = 1+2z+2z^2+2z^3+2z^4+2z^5+2z^6-13z^7$. Since $z^7=1$,this simplifies to $(1-z)S = 1+2(z+z^2+z^3+z^4+z^5+z^6)-13$. Using the property $1+z+z^2+z^3+z^4+z^5+z^6=0$,we have $z+z^2+z^3+z^4+z^5+z^6 = -1$. Thus,$(1-z)S = 1+2(-1)-13 = 1-2-13 = -14$. Therefore,$S = \frac{-14}{1-z}$.

If $z$ is a non-real root of $x^7=1$,then $1+3z+5z^2+7z^3+9z^4+11z^5+13z^6=$

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