Question 14373
 log {{{ (1 + cos x + i* sin x)/(cos x - 1 - i* sin x)  }}}

 Multiplying the complex conjugate of the denominator.

 Since z  = {{{ (1 + cos x + i* sin x)/(cos x - 1 - i* sin x)  }}}
 = {{{ (1 + cos x + i* sin x)* (cos x - 1 + i* sin x)/
  ((cos x - 1 - i* sin x)*(cos x - 1 + i* sin x))  }}}
 = {{{ ( (cos x)^2 -1 - (sin x)^2 + i * sin x *2*cos x )/
   ( (cos x - 1)^2 - (sin x)^2 ) ) }}}
 = {{{ ( -2*(sin x)^2 + i * 2 *sin x *cos x )/
   ( 2*(cos x)^2 - 2* cosx ) ) }}}
 = {{{ (sin x (- sin x + i cos x ))/   (cos x (cos x - 1) ) }}}

 = {{{sin x/(cos x (cos x - 1) ) }}} {{{(- sin x + i cos x) }}}
 = {{{ sin x/(cos x (1-cos x) ) }}} {{{( sin x - i cos x ) }}}
 = {{{sin x/(cos x (1-cos x ) )}}} {{{ (cos(x- pi/2) + i sin (x-pi/2)) }}} 
 (polar coordinates) ...(***)

 The definition of log z for complex number may be beyond your level.

 Anyway,I give it here.
 If the polar form for z = r {{{e^(i*theta) }}}, 

 Here, by (***) the given z =  
(where the radius r = |sin x/(cos x (1-cos x ) )|, 
 the argument {{{theta = x-pi/2 }}})

Hence, the natural logarithm ln(z).
 ln(z) = ln(r) + i * theta = ln|sin x| - ln|cosx (1-cos x) + i {{{(x-pi/2)}}}
 ... Answer
  
 This is a very boring question about calculations, try your best to understand the details.


 Kenny