begin{aligned} & text{If } z=e^{i theta} text | vedclass

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(B) Given $z=e^{i 	heta}$,we have $\cos n 	heta = \frac{z^n+z^{-n}}{2}$ and $\sin n 	heta = \frac{z^n-z^{-n}}{2i}$.Substituting these into the expr

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(B) Given $z=e^{i 	heta}$,we have $\cos n 	heta = \frac{z^n+z^{-n}}{2}$ and $\sin n 	heta = \frac{z^n-z^{-n}}{2i}$.Substituting these into the expression:$\frac{3(\frac{z^3+z^{-3}}{2})+2(\frac{z^2+z^{-2}}{2})+5(\frac{z^5+z^{-5}}{2})}{3(\frac{z^3-z^{-3}}{2i})+2(\frac{z^2-z^{-2}}{2i})+5(\frac{z^5-z^{-5}}{2i})} = i \frac{3z^3+2z^2+5z^5+3z^{-3}+2z^{-2}+5z^{-5}}{3z^3+2z^2+5z^5-3z^{-3}-2z^{-2}-5z^{-5}}$$= i \frac{5z^{10}+3z^8+2z^7+2z^3+3z^2+5}{5z^{10}+3z^8+2z^7-2z^3-3z^2-5} = i \frac{\sum_{r=0}^{10} a_r z^r}{\sum_{r=0}^{10} b_r z^r}$.Comparing the coefficients,we find the sum of coefficients $\sum (a_r+b_r) = 2+3+5 = 10$.Therefore,$\frac{\sum_{r=0}^{10} (a_r+b_r)}{10} = \frac{10}{10} = 1$.

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