Question

Consider a function $f:C\to C$ defined as $f\left(z\right)=z^{12}+2z^{11}+3z^{10}+...+12z+13$ and $S$ be the set defined as $\left\{z|Re\left[f\right(z)-z\left(\frac{z^{13}-1}{(z-1{)}^2}\right)]=\frac{13}{4}\right\},$ where $Re\left(z\right)$ denotes the real part of complex number $z.$ If $\alpha =\cos \frac{2\pi }{13}+i\sin \frac{2\pi }{13},$ where $i=\sqrt{-1},$ then

• A. $f\left(\alpha \right)\cdot f\left({\alpha }^2\right)\cdots f\left({\alpha }^{12}\right)=(13{)}^{11}$
• B. $f\left(\alpha \right)\cdot f\left({\alpha }^2\right)\cdots f\left({\alpha }^{12}\right)=(13{)}^{12}$
• C. the maximum area of the quadrilateral formed by joining four points lying in $S$ is $8.$
• D. the maximum area of the quadrilateral formed by joining four points lying in $S$ is $16.$

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(\alpha \right)\cdot f\left({\alpha }^2\right)\cdots f\left({\alpha }^{12}\right)=(13{)}^{11}$
C the maximum area of the quadrilateral formed by joining four points lying in $S$ is $8.$

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

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.
$\therefore f\left(z\right)=\frac{13}{1-z}(\because z^{13}=1)$
$\Rightarrow f\left(\alpha \right)\cdot f\left({\alpha }^2\right)\cdots f\left({\alpha }^{12}\right)=\frac{13}{1-\alpha }\cdot \frac{13}{1-{\alpha }^2}\cdots \frac{13}{1-{\alpha }^{12}}=\frac{(13{)}^{12}}{(1-\alpha )(1-{\alpha }^2)\cdots (1-{\alpha }^{12})}=\frac{(13{)}^{12}}{13}=(13{)}^{11}$


$f\left(z\right)-\frac{z(z^{13}-1)}{(z-1{)}^2}=\frac{13}{1-z}=\frac{13(1-x+iy)}{(1-x-iy)(1-x+iy)}$
$\therefore Re\left(f\right(z)-\frac{z(z^{13}-1)}{(z-1{)}^2})=\frac{13(1-x)}{(1-x{)}^2+y^2}=\frac{13}{4}$
$\Rightarrow 4-4x=1-2x+x^2+y^2\Rightarrow x^2+y^2+2x-3=0$

$\therefore$ Maximum area of the quadrilateral inscribed in the circle is equal to area of the square $=\frac{d^2}{2}=\frac{16}{2}=8$ square units.