Evaluating the expression is determined by using Euler's theorem
#color(blue)(e^(ix)=cosx+isinx)#
#e^(11pi/6i)=color(blue)(cos(11pi/6)+isin(11pi/6)#
#e^(11pi/6i)=cos((12pi)/6-pi/6)+isin((12pi)/6-pi/6)#
#e^(11pi/6i)=cos(2pi-pi/6)+isin(2pi-pi/6)#
#e^(11pi/6i)=cos(-pi/6)+isin(-pi/6)#
Applying the trigonometric identities:
#color(green)(cos(-alpha)=cosalpha)#
#color(green)(sin(-alpha)=-sinalpha)#
Let us continue computing #e^((11pi)/6)#
#e^(11pi/6i)=cos(-pi/6)+isin(-pi/6)#
#e^(11pi/6i)=color(green)(cos(pi/6))color(green)(-isin(pi/6))#
#e^(11pi/6i)=sqrt3/2-i1/2#
Let us compute #e^((17pi)/12i)#
#e^((17pi)/12i)=color(blue)(cos((17pi)/12)+isin((17pi)/12)#
#e^((17pi)/12i)=cos((12pi)/12+(5pi)/12)+isin((12pi)/12+(5pi)/12)#
#e^((17pi)/12i)=cos(pi+(5pi)/12)+isin(pi+(5pi)/12)#
Applying the trigonometric identities
#color(brown)(cos(pi+alpha)=-cosalpha)#
#color(brown)(sin(pi+alpha)=-sinalpha)#
Let us continue computing #e^((17pi)/12i)#
#e^((17pi)/12i)=cos(pi+(5pi)/12)+isin(pi+(5pi)/12)#
#e^((17pi)/12i)=color(brown)(-cos((5pi)/12)color(brown)-isin((5pi)/12)#
#e^((17pi)/12i)=-(sqrt6-sqrt2)/4-i(sqrt2+sqrt6)/4#
#e^((17pi)/12i)=(-sqrt6+sqrt2)/4-i(sqrt2+sqrt6)/4#
hint:To evaluate #cos((5pi)/12) and sin((5pi)/12)#
#cos((5pi)/12)=cos(pi/6+pi/4)#
# sin((5pi)/12)=sin(pi/6+pi/4)#
So,
#color(blue)(e^(11pi/6i)-e^((17pi)/12i)#
#=sqrt3/2-i1/2-((-sqrt6+sqrt2)/4-i(sqrt2+sqrt6)/4)#
#=sqrt3/2-i1/2-((-sqrt6+sqrt2)/4)+i(sqrt2+sqrt6)/4#
#=sqrt3/2-i1/2+sqrt6/4-sqrt2/4+i(sqrt2+sqrt6)/4#
#=(2sqrt3)/4-(2i)/4+sqrt6/4-sqrt2/4+i(sqrt2+sqrt6)/4#
#=(2sqrt3)/4+sqrt6/4-sqrt2/4+i(sqrt2+sqrt6)/4-(2i)/4#
#=(2sqrt3+sqrt6-sqrt2)/4+i(sqrt2+sqrt6-2)/4#
