1. ## complex numbers factorisation

how do you find k for
l 1- exp (ix) l = k l sin ( x/2) l for all real x

i tried to square both sides of the equation and i got:

1- ( cos (^2) x + sin(^2) x ) -2i sin x( 1- cos x) -2 cos x = k^2 sin(^ 2) ( x/2)

factorising the sin (^2) ( x/2) terms to one side,
sin (^2) (x/2) ( 4 - 4 i sin x) -2 = k(^2) sin(^2) ( x/2)

and then i got stuck..

2. $|1 - e^{ix} |= |e^{\frac{ix}{2}} ( e^{-\frac{ix}{2}} - e^{\frac{ix}{2}} ) |$

$= |e^{\frac{ix}{2}} | |e^{-\frac{ix}{2}} - e^{\frac{ix}{2}} |$

$= (1) |-2i \sin(\frac{x}{2})| = |-2i| |\sin(\frac{x}{2})|$

$= 2 |\sin(\frac{x}{2})|$

By comparing the multiple , $k = 2$