"Reminder : "
Euler's relation
e^(itheta)=costheta+isintheta
Here, we have
z=e^(13/8pi)-e^(7/4pi)=cos(13/8pi)+isin(13/8pi)-cos(7/4pi)-isin(7/4pi)
Therefore,
13/8pi=5/8pi+pi=1/8pi+3/2pi
7/4pi=3/4pi+pi
cos(1/4pi)=1-2sin^2(1/8pi)=2cos^2(1/8pi)-1
sin(1/8pi)=sqrt((1-cos(1/4pi))/2)=sqrt((1-sqrt2/2)/(2))=(sqrt(2-sqrt2))/2
cos(1/8pi)=sqrt((1+cos(1/4pi))/2)=sqrt((1+sqrt2/2)/(2))=(sqrt(2+sqrt2))/2#
So,
z=cos(1/8pi+3/2pi)+isin(1/8pi+3/2pi)-cos(1/4pi+pi)-isin(1/4pi+pi)
cos(1/8pi+3/2pi)=cos(1/8pi)cos(3/2pi)-sin(1/8pi)sin(3/2pi)
=(sqrt(2+sqrt2))/2*0-(sqrt(2-sqrt2))/2*(-1)=(sqrt(2-sqrt2))/2
sin(1/8pi+3/2pi)=sin(1/8pi)cos(3/2pi)+cos(1/8pi)sin(3/2pi)
=(sqrt(2-sqrt2))/2*0+(sqrt(2+sqrt2))/2*(-1)=-(sqrt(2+sqrt2))/2
cos(3/4pi+pi)=cos(3/4pi)cos(pi)-sin(3/4pi)sin(pi)
=-sqrt2/2*-1-sqrt2/2*0=sqrt2/2
sin(3/4pi+pi)=sin(3/4pi)cos(pi)+cos(3/4pi)sin(pi)
=sqrt2/2*-1+sqrt2/2*0=-sqrt2/2
Finally,
z=(sqrt(2-sqrt2))/2-i(sqrt(2+sqrt2))/2-sqrt2/2+isqrt2/2
=((sqrt(2-sqrt2))/2-sqrt2/2)-i((sqrt(2+sqrt2))/2-sqrt2/2)