Question

If
$z=e^{i \theta}$ and $\frac{3 \cos 3 \theta+2 \cos 2 \theta+5 \cos 5 \theta}{3 \sin 3 \theta+2 \sin 2 \theta+5 \sin 5 \theta}=\frac{i \displaystyle\sum_{r=0}^{10} a_{r} z^{r}}{\displaystyle\sum_{r=0}^{10} b_{r} z^{r}}$ then
$\frac{\left(\displaystyle\sum_{r=0}^{10} a_{r}+\displaystyle\sum_{r=0}^{10} b_{r}\right)}{10}=$

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

For a complex number $Z=e^{i \theta}$ it is given that
$\frac{3 \cos 3 \theta+2 \cos 2 \theta+5 \cos 5 \theta}{3 \sin 3 \theta+2 \sin 2 \theta+5 \sin 5 \theta}=i \frac{\displaystyle\sum_{r=0}^{10} a_{r} Z^{r}}{\displaystyle\sum_{r=0}^{10} b_{r} Z^{r}}$
$\Rightarrow \frac{\displaystyle\sum_{r=0}^{10} a_{r} z^{r}+\displaystyle\sum_{r=0}^{10} b_{r} z^{r}}{\displaystyle\sum_{r=0}^{10} a_{r} z^{r}-\displaystyle\sum_{r=0}^{10} b_{r} z^{r}}=\frac{3 e^{i 3 \theta}+2 e^{i 2 \theta}+5 e^{i 5 \theta}}{3 e^{-i 3 \theta}+2 e^{-i 2 \theta}+5 e^{-i 5 \theta}} $
$\Rightarrow \frac{\displaystyle\sum_{r=0}^{10} z^{r}\left(a_{r}+b_{r}\right)}{\displaystyle\sum_{r=0}^{10} z^{r}\left(a_{r}-b_{r}\right)}=\frac{3 e^{i 3 \theta}+2 e^{i 2 \theta}+5 e^{i 5 \theta}}{3 e^{-i 3 \theta}+2 e^{-i 2 \theta}+5 e^{-i 5 \theta}} $
$\Rightarrow a_{0}+b_{0}=a_{1}+b_{1}=a_{4}+b_{4}=a_{6}+b_{6}$
$=a_{7}+b_{7}=a_{8}+b_{8}$
$=a_{9}+b_{9}=a_{10}+b_{10}=0$
and
$a_{2}+b_{2}=2, a_{3}+b_{3}=3, a_{5}+b_{5}=5$
$\therefore \frac{\displaystyle\sum_{r=0}^{10}\left(a_{r}+b_{r}\right)}{10}=\frac{2+3+5}{10}=1$