, then 
is equal to
A)
B)
C)
D)
Correct Answer: C
Solution :
Let \[z=\frac{-1}{2}+\frac{i\,\sqrt{3}}{2}=\omega \] \[8+10\,z+7{{z}^{2}}=8+10\omega +7{{\omega }^{2}}\] \[=\,\,\,1+3\,\omega +7+7\,\omega +7\,{{\omega }^{2}}\] \[=\,\,1+3\,\omega +7(1+\omega +{{\omega }^{2}})\] \[=\,\,1+3\omega \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because \,\,\,\,1+\omega +{{\omega }^{2}}=0)\] \[1+3\left( -\frac{1}{2}+\frac{i\sqrt{3}}{2} \right)=\frac{-1+3\sqrt{3}\,i}{2}\]
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