We apply Euler's relation
#e^(itheta)=costheta+isintheta#
#e^(1/12pii)=cos(1/12pi)+isin(1/12pi)#
#e^(9/8pii)=cos(9/8pi)+isin(9/8pi)#
#cos(1/12pi)=cos(1/3pi-1/4pi)#
#=cos(1/3pi)cos(1/4pi)+sin(1/3pi)sin(1/4pi)#
#=1/2*sqrt2/2+sqrt3/2*sqrt2/2#
#=(sqrt2+sqrt6)/4#
#sin(1/12pi)=sin(1/3pi-1/4pi)#
#=sin(1/3pi)cos(1/4pi)-sin(1/4pi)cos(1/3pi)#
#=sqrt3/2*sqrt2/2-sqrt2/2*1/2#
#=(sqrt6-sqrt2)/4#
#cos(9/8pi)=cos(pi+1/8pi)#
#=cos(pi)cos(1/8pi)-sin(pi)cos(1/8pi)#
#=-1*cos(1/8pi)-0*cos(1/8pi)#
#=-cos(1/8pi)#
#Cos(2x)=2cos^2x-1#
#cos^2x=(1+cos(2x))/2#
#cos^2(pi/8)=1/2(1+cos(pi/4))=1/2(1+sqrt2/2)=(2+sqrt2)/4#
#cos(pi/8)=sqrt(2+sqrt2)/2#
#cos(9/8pi)=-sqrt(2+sqrt2)/2#
#sin(9/8pi)=sin(pi+1/8pi)=sin(pi)cos(1/8pi)+sin(1/8)picos(pi)#
#=0-sin(1/8pi)#
#cos^2(1/8pi)+sin^2(1/8pi)=1#
#sin^2(1/8pi)=1-(sqrt(2+sqrt2)/2)^2=(4-2-sqrt2)/4#
#sin(1/8pi)=sqrt(2-sqrt2)/2#
#sin(9/8pi)=-sqrt(2-sqrt2)/2#
Therefore,
#e^(1/12pii)-e^(9/8pii)=cos(1/12pi)+isin(1/12pi)-cos(9/8pi)+isin(9/8pi)#
#=(sqrt2+sqrt6)/4+sqrt(2+sqrt2)/2+i((sqrt6-sqrt2)/4+sqrt(2-sqrt2)/2)#
