Question

For any integer $k$, if $a_k=\cos \left(\frac{k\pi }{7}\right)+i\sin \left(\frac{k\pi }{7}\right)$, then value of the expression $\frac{\sum _{k=1}^{12}|a_{k+1}-a_k|}{\sum _{k=1}^3|a_{4k-1}-a_{4k-2}|}$ is

Answer 1

$a_k=\cos \left(\frac{k\pi }{7}\right)+i\sin \left(\frac{k\pi }{7}\right)=e^{i\left(\frac{k\pi }{7}\right)}$
So,
$a_{k+1}-a_k=e^{i\left(\frac{(k+1)\pi }{7}\right)}-e^{i\left(\frac{k\pi }{7}\right)}=e^{i\left(\frac{k\pi }{7}\right)}(e^{\frac{i\pi }{7}}-1)=e^{i\left(\frac{k\pi }{7}\right)}{e}^{\frac{i\pi }{14}}(e^{\frac{i\pi }{14}}-e^{\frac{-i\pi }{14}})=e^{i\left(\frac{(2k+1)\pi }{7}\right)}\times \left[2i\sin \left(\frac{\pi }{14}\right)\right]\Rightarrow |a_{k+1}-a_k|=2\sin \left(\frac{\pi }{14}\right)$

Replacing $k$ with $4k-2$ in the above expression,
$|a_{4k-1}-a_{4k-2}|=2\sin \left(\frac{\pi }{14}\right)$
Hence the value of the expression,
$\frac{\sum _{k=1}^{12}|a_{(k+1)}-a_k|}{\sum _{k=1}^3|a_{(4k-1)}-a_{(4k-2)}|}=\frac{24}{6}=4$


Alternate Solution:
$a_k=\cos \left(\frac{k\pi }{7}\right)+i\sin \left(\frac{k\pi }{7}\right)$
Then
$|a_{k+1}-a_k|=|(\cos \frac{(k+1)\pi }{7}-\cos \frac{k\pi }{7})+i(\sin \frac{(k+1)\pi }{7}-\sin \frac{k\pi }{7})|=\sqrt{2-2(\cos \frac{(k+1)\pi }{7}\cos \frac{k\pi }{7}+\sin \frac{(k+1)\pi }{7}\sin \frac{k\pi }{7})}=\sqrt{2-2\cos \frac{\pi }{7}}=\sqrt{2(1-\cos \frac{\pi }{7})}\Rightarrow |a_{k+1}-a_k|=2\sin \frac{\pi }{14}$
Replace $k$ with $4k-2$ in above equation
$\Rightarrow |a_{4k-2}-a_{4k-2}|=2\sin \frac{\pi }{14}$
Hence the value of the expression,
$\frac{\sum _{k=1}^{12}|a_{(k+1)}-a_k|}{\sum _{k=1}^3|a_{(4k-1)}-a_{(4k-2)}|}=\frac{24}{6}=4$