The expression is evaluated by applying Euler's formula:
color(red)(e^(ix)=cosx+isinx)eix=cosx+isinx
We will use the following trigonometric identities:
color(blue)(cos(2pi+alpha)=cosalphacos(2π+α)=cosα
color(blue)(cos(-alpha)=cosalpha)cos(−α)=cosα
color(blue)(sin(-alpha)=-sinalpha)sin(−α)=−sinα
Let us compute e^(pi/2i)and e^((5pi)/3i)eπ2iande5π3i by applying the above formula
e^(pi/2i)=cos(pi/2)+isin(pi/2)eπ2i=cos(π2)+isin(π2)
rArre^(pi/2i)=0+i(1)⇒eπ2i=0+i(1)
rArre^(pi/2i)=i⇒eπ2i=i
e^((5pi)/3i)=cos((5pi)/3)+isin((5pi)/3)e5π3i=cos(5π3)+isin(5π3)
rArre^((5pi)/3i)=cos((6pi)/3-pi/3)+isin((6pi)/3-pi/3)⇒e5π3i=cos(6π3−π3)+isin(6π3−π3)
rArre^((5pi)/3i)=cos(2pi-pi/3)+isin(2pi-pi/3)⇒e5π3i=cos(2π−π3)+isin(2π−π3)
rArre^((5pi)/3i)=color(blue)(cos(-pi/3)+isin(-pi/3))⇒e5π3i=cos(−π3)+isin(−π3)
rArre^((5pi)/3i)=color(blue)(cos(pi/3)-isin(pi/3)⇒e5π3i=cos(π3)−isin(π3)
rArre^((5pi)/3i)=1/2-isqrt3/2⇒e5π3i=12−i√32
e^(pi/2i)-e^((5pi)/3i)eπ2i−e5π3i
=i-(1/2-isqrt3/2)=i−(12−i√32)
=i-1/2+isqrt3/2=i−12+i√32
=-1/2+i(1+sqrt3/2)=−12+i(1+√32)
Therefore,
e^(pi/2i)-e^((5pi)/3i)=-1/2+i(1+sqrt3/2)eπ2i−e5π3i=−12+i(1+√32)
