Question

Let $z$ be any complex number. To factorise the expression of the form ($z^n-1$), we consider the equation $z^n=1$. This equation is solved using De moiver's theorem. Let $1,{\alpha }_1,{\alpha }_2.......{\alpha }_{n-1}$ be the roots of this equation, then $z^n-1=(z-1)(z-{\alpha }_1)(z-{\alpha }_2).......(z-{\alpha }_{n-1})$. This method can be generalised to factorize any expression of the form $z^n-k^n$.

For example, $z^7+1={\Pi }_{m=0}^6(z-CiS(\frac{2m\pi }{7}+\frac{\pi }{7}))$

This can be further simplified as

$z^7+1(z+1)(z^2-2zcos\frac{\pi }{7}+1)(z^2-2zcos\frac{3\pi }{7}+1)(z^2-2zcos\frac{5\pi }{7}+1)$ .......$\left(i\right)$

These factorisations are useful in proving different trigonometric identities e.g. in equation $\left(i\right)$ if we put $z=i,$ then equation $\left(i\right)$ becomes

$(1-i)=(i+1)(-2icos\frac{\pi }{7})(-2icos\frac{3\pi }{7})(-2icos\frac{5\pi }{7})$

i.e, $cos\frac{\pi }{7}cos\frac{3\pi }{7}cos\frac{5\pi }{7}=-\frac{1}{8}$

By using the factorisation for $z^5+1$, the value of $4sin\frac{\pi }{10}cos\frac{\pi }{5}$ comes out to be :

• A. $4$
• B. $1/4$
• C. $1$
• D. $-1$

Answer 1

Answer: (C)

$1$

$z^5+1=0$ Implies

$z=-1,(cos\frac{\pi }{5}+isin\frac{\pi }{5}),(cos\frac{-\pi }{5}+isin\frac{-\pi }{5}),(cos\frac{3\pi }{5}+isin\frac{3\pi }{5}),(cos\frac{-3\pi }{5}+isin\frac{-3\pi }{5})$

Now

$4sin\frac{\pi }{10}.cos\frac{\pi }{5}$
$=4sin\frac{\pi }{10}sin\frac{3\pi }{10}$

$=2[cos\frac{\pi }{5}-cos\frac{2\pi }{5}]$

$=2[cos\frac{\pi }{5}+cos\frac{3\pi }{5}]$

$=cos\frac{\pi }{5}+cos\frac{\pi }{5}+cos\frac{3\pi }{5}+cos\frac{3\pi }{5}$

$=cos\frac{\pi }{5}+cos\frac{3\pi }{5}+cos\frac{-3\pi }{5}+cos\frac{-\pi }{5}$

$=\sum Re{z}_i$

$=|z|$
$=1$