Question

Consider a function $f:C\to C$ defined as $f\left(z\right)=z^{12}+2z^{11}+3z^{10}+...+12z+13$. If $\alpha =\cos \frac{2\pi }{13}+i\sin \frac{2\pi }{13},$ where $i=\sqrt{-1},$ then the value of $f\left(\alpha \right)f\left({\alpha }^2\right)f\left({\alpha }^3\right)\dots f\left({\alpha }^{12}\right)$ is

• A. ${13}^{10}$
• B. ${13}^{13}$
• C. ${13}^{12}$
• D. ${13}^{11}$

Answer 1

The correct option is D ${13}^{11}$

$f\left(z\right)=z^{12}+2\cdot z^{11}+3\cdot z^{10}+...+12\cdot z+13$
$\underset{–}{f\left(z\right)\cdot \frac{1}{z}=z^{11}+2\cdot z^{10}+...+11z+12+\frac{13}{z}}$
$f\left(z\right)(1-\frac{1}{z})=z^{12}+z^{11}+...+1-\frac{13}{z}$
$\Rightarrow f\left(z\right)\left(\frac{z-1}{z}\right)=\frac{z^{13}-1}{z-1}-\frac{13}{z}$
$\Rightarrow f\left(z\right)=\frac{z(z^{13}-1)}{(z-1{)}^2}-\frac{13}{z-1}$

For $\alpha =\cos \frac{2\pi }{13}+i\sin \frac{2\pi }{13},$
$1,\alpha ,{\alpha }^2,...,{\alpha }^{12}$ are the $13$ roots of unity.
$\therefore f\left(z\right)=\frac{13}{1-z}(\because z^{13}=1)$
$\Rightarrow f\left(\alpha \right)\cdot f\left({\alpha }^2\right)\cdots f\left({\alpha }^{12}\right)=\frac{13}{1-\alpha }\cdot \frac{13}{1-{\alpha }^2}\cdots \frac{13}{1-{\alpha }^{12}}=\frac{(13{)}^{12}}{(1-\alpha )(1-{\alpha }^2)\cdots (1-{\alpha }^{12})}$

As $z^{13}=1$
$\Rightarrow \underset{z\to 1}{lim}(z-\alpha )(z-{\alpha }^2)\cdots (z-{\alpha }^{12})=\underset{z\to 1}{lim}\frac{z^{13}-1}{z-1}\Rightarrow \underset{z\to 1}{lim}(z-\alpha )(z-{\alpha }^2)\cdots (z-{\alpha }^{12})=13$
Hence,
$f\left(\alpha \right)f\left({\alpha }^2\right)f\left({\alpha }^3\right)\dots f\left({\alpha }^{12}\right)={13}^{11}$