Euler's Identity states that #e^(ix)=cosx+isinx#, and from that #e^((23pi)/12i)=cos((23pi)/12)+isin((23pi)/12)# and #e^((13pi)/2i)=cos((13pi)/2)+isin((13pi)/2)#.
Substituting these into your problem, we have #cos((23pi)/12)+isin((23pi)/12)-cos((13pi)/2)-isin((13pi)/2)#. If you can use a calculator then you can stop here to find the values of these functions.
If you need an exact answer there's a bit farther to go in order to put these functions in terms of special angles.
Since #sin(theta)=sin(theta+2npi)# and #cos(theta)=cos(theta+2npi)# where #ninZZ# we can say #n="-"6# and change our expression to #cos((23pi)/12)+isin((23pi)/12)-cos((pi)/2)-isin((pi)/2)#
Since #sin(2npi-theta)=sin("-"theta)="-"sin(theta)# and #cos(2npi-theta)=cos("-"theta)=cos(theta)# where #ninZZ# we can say #n=1# and change our expression to #cos((pi)/12)-isin((pi)/12)-cos((pi)/2)-isin((pi)/2)#.
Finally, since #pi/12=(pi/6)/2#, our expression is #cos((pi/6)/2)-isin((pi/6)/2)-cos((pi)/2)-isin((pi)/2)# and we can use the half-angle formulas #sin(theta/2)=+-sqrt((1-cos(theta))/2)# and #cos(theta/2)=+-sqrt((1+cos(theta))/2)# leaving us with #+-sqrt((1+cos(pi/6))/2)-i(+-sqrt((1-cos(pi/6))/2))-cos((pi)/2)-isin((pi)/2)#. Ugly, I know, but easy since #cos(pi/6)=sqrt3/2#, #cos(pi/2)=0#, and #sin(pi/2)=1#.
#"+"sqrt((1+sqrt3/2)/2)-i("+"sqrt((1-sqrt3/2)/2))-0-i# so the answer is #sqrt((1+sqrt3/2)/2)-isqrt((1-sqrt3/2)/2)-i#
