Tag: complex numbers

The roots of the equation $x^{2n + 1}- 1 = 0$ are
$1, \displaystyle cos \frac{2 \pi}{2n + 1} + i  sin  \frac{2 \pi}{2n + 1}, cos \frac{4 \pi}{2n + 1} + i  sin \frac{4 \pi}{2n + 1}, ........, cos \frac{4 n \pi}{2n + 1} + i  sin  \frac{4n  \pi}{2n + 1}$
Therefore $\displaystyle x^{2n+1} - 1 = (x - 1) \left ( x - cos \frac{2 \pi}{2n + 1} - i  sin \frac{2 \pi}{2n + 1} \right ) \left ( x - cos \frac{4 \pi}{2n + 1} - i  sin \frac{4 \pi}{2n + 1} \right ) ....... \left ( x - cos \frac{4 n\pi}{2n + 1} - i  sin \frac{4n \pi}{2n + 1} \right )$
Further since
$\displaystyle cos \left ( \frac{(2n + 1) - r}{2n + 1} \right ) 2\pi = cos \frac{2 r \pi}{2n + 1}$
and $\displaystyle sin \left ( \frac{(2n + 1) - r}{2n + 1} \right ) 2\pi = -sin \frac{2 r \pi}{2n + 1}$
it follows that
$\displaystyle \left ( x - cos \frac{2 \pi}{2n + 1} - i  sin \frac{2 \pi}{2n + 1}\right ) \left ( x - cos \frac{4 \pi}{2n + 1} - i  sin \frac{4 \pi}{2n + 1}\right )$
$= x^2 - 2x  cos \displaystyle \frac{2 \pi}{2n + 1} + 1$
$\left ( x - cos \frac{4 \pi}{2n + 1} - i  sin \frac{4 \pi}{2n + 1}\right )\left ( x - cos \frac{(4n - 2) \pi}{2n + 1} - i  sin \frac{(4n - 2) \pi}{2n + 1}\right )$
$=x^2 - 2x  cos \displaystyle \frac{4 \pi}{2n + 1} + 1$
$\left ( x - cos \frac{2 n\pi}{2n + 1} - i  sin \frac{2 n\pi}{2n + 1}\right )\left ( x - cos \frac{(2n + 2) \pi}{2n + 1} - i  sin \frac{(2n + 2)}{(2n + 1)} \pi\right )$
$= x^2 - 2x   cos \frac{2 n \pi}{2n + 1} + 1$
Thus the polynomial $x^{2n + 1} - 1$ can be rewritten thus
$x^{2n + 1} - 1 = (x - 1) \displaystyle \left ( x^2 - 2x  cos  \frac{2 \pi}{2n + 1} + 1\right ) \left ( x^2 - 2x  cos  \frac{4 \pi}{2n + 1} + 1\right )........ \left ( x^2 - 2x  cos  \frac{2 n\pi}{2n + 1} + 1\right ) $
or $\displaystyle \frac{x^{2n + 1} - 1}{x - 1} = \left ( x^2 - 2x  cos  \frac{2 \pi}{2n + 1} + 1\right ) \left ( x^2 - 2x  cos  \frac{4 \pi}{2n + 1} + 1\right ) ......... \left ( x^2 - 2x  cos  \frac{2 n\pi}{2n + 1} + 1\right ) $
Taking $\displaystyle \lim _{x \rightarrow 1}$ on both sides
$(2n + 1) = 2^{2n} sin^2 \displaystyle \frac{\pi}{2n + 1} sin^2 \frac{2 \pi}{2n + 1} ..... sin^2 \frac{n \pi}{2n + 1}$
Hence, $\displaystyle sin \frac{\pi}{2n + 1} sin \frac{2 \pi}{2n + 1} ...... sin \frac{n \pi}{2n + 1} = \frac{\sqrt{(2n + 1)}}{2^n}$

$1,\alpha ,{ \alpha  }^{ 2 }...{ \alpha  }^{ n-1 }$ are the nth roots of unity. Thus, they are solutions of the equation: ${ x }^{ n }-1=0$. 
Thus, product of roots $= { (-1) }^{ n }(\dfrac { -1 }{ 1 } )$
(i.e. ${ (-1) }^{ n }$constant term / coefficient of ${ x }^{ n }$)
Thus, the product = ${ (-1) }^{ n }(\dfrac { -1 }{ 1 } )={ (-1) }^{ n }(\dfrac { -1 }{ 1 } )={ (-1) }^{ n+1 }={ (-1) }^{ 2 }{ (-1) }^{ n-1 }=1{ (-1) }^{ n-1 }={ (-1) }^{ n-1 }$
Hence, (A) is correct.

Given : $\displaystyle \alpha = \cos\dfrac{8\pi}{11}+i\sin\dfrac{8\pi }{11}$
$\displaystyle \therefore \alpha ^{11}= \cos 8\pi +i\sin 8\pi = 1$
$\displaystyle \sum _{n= 0}^{10}\alpha ^{n}= \dfrac{1-\alpha ^{11}}{1-\alpha }= 0$ (sum of 11th roots of unity)
Now $\displaystyle Re(z)= \dfrac{z+\bar{z}}{2}=\dfrac{sum \ of\  11 \ roots\  of\  unity -1}{2}= -1/2 $

$S=1+2\alpha+3\alpha^{2}+...n\alpha^{n-1}$
Here $S$ forms an A.G.P
Therefore
$S=1+2\alpha+3\alpha^{2}+...n\alpha^{n-1}$
$S(\alpha)=\alpha+2\alpha^{2}+3\alpha^{3}+...(n-1)\alpha^{n-1}+n\alpha^{n}$
Hence
$S(1-\alpha)=1+\alpha+\alpha^{2}+....\alpha^{n-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{1-\alpha^{n}}{1-\alpha}-n\alpha^{n}$
$S(1-\alpha)=0-n$
$S=\dfrac{-n}{1-\alpha}$
Hence, option 'A' is correct.

$\displaystyle \frac{2^{2n + 1} - 1}{x - 1} = \left ( x^2 - 2x   cos  \frac{2 \pi}{2n + 1} + 1\right ) \left ( x^2 - 2x  cos \frac{4 \pi}{2n + 1} + 1\right ) ....... \left ( x^2 - 2x   cos \frac{2n  \pi}{2n + 1}
 + 1\right )$
Taking $\displaystyle \lim _{x \rightarrow - 1}$ on both sides, we have
$1 = 4^n cos^2 \displaystyle \frac{\pi}{2n + 1} cos^2 \frac{2 \pi}{2n + 1} ..... cos^2 \frac{n \pi}{2n + 1}$
$\therefore \displaystyle cos \frac{\pi}{2n + 1} cos \frac{2\pi}{2n + 1} ..... cos \frac{n \pi}{2n + 1} = \frac{1}{2^n}$

$\left| 1+r+{ r }^{ 2 }+{ r }^{ -2 }-{ r }^{ -1 } \right| =\left| 1+r+{ r }^{ 2 }+{ r }^{ 3 }-{ r }^{ 4 } \right| $

$z^{5}+1=0$ Implies 

$\displaystyle 2\sum ab= \left ( \sum a \right )^{2}-\sum a^{2}$
$\displaystyle \therefore 2\sum \sum \left ( \alpha _{i}\alpha _{j} \right )^{5}= \left ( \alpha _{1}^{5}+\alpha _{2}^{5}+\cdots  \right )^{2}-\left ( \alpha _{1}^{10}+\alpha _{2}^{10}+\cdots  \right )$
$\displaystyle = 0-0$ ($\displaystyle \because \sum \alpha _{i}^{r}= 100$ if $\displaystyle r= 100k$
and $ \sum \alpha _{i}^{r} = 0$ if $\displaystyle r\neq 100k$)
Here both 5 and 10 are not multiples of 100.

$\displaystyle \sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7}$
$\displaystyle = -i^{2}\sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7}$
$\displaystyle = -i\left ( \cos \frac{2\pi k}{7}+i\sin \frac{2\pi k}{7} \right )= -i e^{i2\pi k/7}= -iz^{k}$
where, $\displaystyle z= \cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7}$ or $\displaystyle z^{7}= 1$
or $\displaystyle z= \left ( 1 \right )^{1/7} \therefore 1+z+2^{2}+\cdots +z^{6}= 0$ ...(1)
Above being the sum of seven, seventh roots of unity
$\displaystyle \sum = -i \sum _{i= 1}^{k} z^{k}= -1\left ( z+z^{2}+\cdots +z^{6} \right )= -i\left ( -1 \right )= i$ by (I)
Alternative Method. $\displaystyle \sin \theta -i\cos \theta $
$\displaystyle = -i^{2}\sin \theta -i\cos \theta = -i\left ( \cos \theta +i\sin \theta  \right )$
$\displaystyle = -ie^{i\theta }$ where $\displaystyle \theta = \frac{2\pi }{7}$ or $\displaystyle 7\theta = 2\pi $
Hence the given sigma is
$\displaystyle \sum _{k= 1}^{6}-ie^{ik\theta }= -i\left [ e^{i\theta +}e^{2i\theta }+\cdots +e^{6i\theta } \right ]$
$\displaystyle = -ie^{i\theta }\left [ \frac{1-\left ( e^{i\theta } \right )^{6}}{1-e^{i\theta }} \right ]= -i\left [ \frac{e^{i\theta }-e^{7i\theta }}{1-e^{i\theta }} \right ]$ (G.P.)
$\displaystyle = -i\left [ \frac{e^{i\theta }-1}{1-e^{i\theta }} \right ]= -i\left ( -1 \right )= i$
$\displaystyle \because 7i\theta = 2\pi i \therefore e^{7i\theta }= \left ( \cos 2\pi +i\sin 2\pi  \right )= 1.$

${ (cosx+i sinx) }^{ n }=cos(nx)+i sin(nx)$
Comparing imaginary parts: $sin(nx)= _{ 1 }^{ n }{ C }sinx{ .cos }^{ n-1 }x- _{ 3 }^{ n }{ C }sin^{ 3 }x{ .cos }^{ n-3 }x+.....$
Divide both sides by $sin^{ n }x$
$\dfrac { sin(nx) }{ sin^{ n }x } = _{ 1 }^{ n }{ C }cot^{ n-1 }x- _{ 3 }^{ n }{ C }cot^{ n-3 }x+...$
Replace $n$ with $2n+1:$ $\dfrac { sin((2n+1)x) }{ sin^{ 2n+1 }x } = _{ 1 }^{ 2n+1 }{ C }cot^{ 2n }x- _{ 3 }^{ 2n+1 }{ C }cot^{ 2n-2 }x+...$
$ = _{ 1 }^{ 2n+1 }{ C }(cot^{ 2 }x)^{ n }- _{ 3 }^{ 2n+1 }{ C }(cot^{ 2 }x)^{ n-1 }+.... ----1)$
So $cot^{ 2 }x _{ i }$ are the roots of the polynomial for LHS=0 where $cot^{ 2 }x _{ i }=cot^{ 2 }(\dfrac { \pi i }{ 2n+1 } )$ where i=1,2,3,...n.
Sum of the roots is the ratio of the first two coefficients on the RHS.
Thus $\sum _{ i=1 }^{ n }{ cot^{ 2 }(\dfrac { \pi i }{ 2n+1 } )= } -\dfrac { - _{ 3 }^{ 2n+1 }{ C } }{ _{ 1 }^{ 2n+1 }{ C } } =\dfrac { (2n+1)2n(2n-1) }{ 6 } .\dfrac { 1 }{ 2n+1 } =\dfrac { n(2n-1) }{ 3 } $
Hence, (c) is correct.