Apply Euler's Identity
e^(itheta)=costheta+isinthetaeiθ=cosθ+isinθ
i^2=-1i2=−1
Therefore,
e^(ipi/2)=cos(pi/2)+isin(pi/2)=0+i*1=ieiπ2=cos(π2)+isin(π2)=0+i⋅1=i
e^(i19/12pi)=cos(19/12pi)+isin(19/12pi)ei1912π=cos(1912π)+isin(1912π)
cos(19/12pi)=cos(5/4pi+1/3pi)cos(1912π)=cos(54π+13π)
=cos(5/4pi)cos(1/3pi)-sin(5/4pi)sin(1/3pi)=cos(54π)cos(13π)−sin(54π)sin(13π)
=(-sqrt2/2)*(1/2)-(-sqrt2/2)*(sqrt3/2)=(−√22)⋅(12)−(−√22)⋅(√32)
=-sqrt2/4+sqrt6/4=−√24+√64
=(sqrt6-sqrt2)/4=√6−√24
sin(19/12pi)=sin(5/4pi+1/3pi)sin(1912π)=sin(54π+13π)
=sin(5/4pi)cos(1/3pi)+cos(5/4pi)sin(1/3pi)=sin(54π)cos(13π)+cos(54π)sin(13π)
=(-sqrt2/2)*(1/2)+(-sqrt2/2)*(sqrt3/2)=(−√22)⋅(12)+(−√22)⋅(√32)
=-sqrt2/4-sqrt6/4=−√24−√64
=-(sqrt6+sqrt2)/4=−√6+√24
Finally,
e^(i19/12pi)*e^(ipi/2)=((sqrt6-sqrt2)/4-i(sqrt6+sqrt2)/4)*(i)ei1912π⋅eiπ2=(√6−√24−i√6+√24)⋅(i)
=(sqrt6+sqrt2)/4+i(sqrt6-sqrt2)/4=√6+√24+i√6−√24
