We apply Euler's formula
#e^(itheta)=costheta+isintheta#
#e^(i5/12pi)=cos(5/12pi)+isin(5/12pi)#
#cos(5/12pi)=cos(3/12pi+2/12pi)#
#=cos(1/4pi+1/6pi)#
#=cos(1/4pi)cos(1/6pi)-sin(1/4pi)sin(1/6pi)#
#=sqrt2/2*sqrt3/2-sqrt2/2*1/2#
#=(sqrt6-sqrt2)/4#
#sin(5/12pi)=sin(1/4pi+1/6pi)#
#=sin(1/4pi)cos(1/6pi)+cos(1/4pi)sin(1/6pi)#
#=sqrt2/2*sqrt3/2+sqrt2/2*1/2#
#=(sqrt6+sqrt2)/4#
Therefore,
#e^(i5/12pi)=(sqrt6-sqrt2)/4+i(sqrt6+sqrt2)/4#
#e^(i3/2pi)=cos(3/2pi)+isin(3/2pi)#
#=0-i#
So,
#e^(i5/12pi).e^(i3/2pi)=((sqrt6-sqrt2)/4+i(sqrt6+sqrt2)/4)*-i#
#=(sqrt6+sqrt2)/4-i(sqrt6-sqrt2)/4#
As #i^2=-1#
Verification
#e^(i5/12pi)*e^(i3/2pi)=e^(i(5/12+3/2)pi)#
#=e^(i23/12pi)#
#=cos(-1/12pi)+isin(-1/12pi)#
#cos(-1/12pi)=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#
#=(sqrt6+sqrt2)/4#
#sin(-1/12pi)=-sin(1/3pi-1/4pi)#
#=-sin(1/3pi)cos(1/4pi)-cos(1/3pi)sin(1/4pi)#
#=-(sqrt3/2*sqrt2/2-1/2*sqrt2/2)#
#=-(sqrt6-sqrt2)/4#
Therefore,
#e^(i23/12pi)=(sqrt6+sqrt2)/4-i(sqrt6-sqrt2)/4#
The result is the same.
