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.
