Question

The sum $\sum _{k=1}^6(\sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7})$ simplyfies to a pure imaginary number. Find its modulus.

Answer 1

Since, $cos\theta +isin\theta =e^{i\theta }$

$z=\sum _{k=1}^6[\sin \frac{2k\pi }{7}-i\cos \frac{2k\pi }{7}]=-i\sum _{k=1}^6\left[cos\right(\frac{2k\pi }{7})+isin(\frac{2k\pi }{7}\left)\right]=-i\sum _{k=1}^6{e}^{i\frac{2k\pi }{7}}$

Solving the index part only which is

$i\frac{2k\pi }{7}=i\frac{2\pi }{7}(1+2+3+\cdots +6)$......[putting the values of $k$]

$=i6\pi$

So,

$z=-i{\sum }_{k=1}^6{e}^{i\frac{2k\pi }{7}}=-i{e}^{i6\pi }=-i(cos6\pi +sin6\pi )=-i$

$z=-i$

Hence $|z|=1$