e^( ( 5 pi)/4 i) - e^( ( 19 pi)/12i)e5π4i−e19π12i
=[cos((5pi)/4)+isin((5pi)/4)] -[cos((19pi)/12)+isin((19pi)/12)]=[cos(5π4)+isin(5π4)]−[cos(19π12)+isin(19π12)]
=[cos((5pi)/4)-cos((19pi)/12)]-i[sin((19pi)/12)-sin((5pi)/4)]=[cos(5π4)−cos(19π12)]−i[sin(19π12)−sin(5π4)]
=[cos((15pi)/12)-cos((19pi)/12)]-i[sin((19pi)/12)-sin((15pi)/12)]=[cos(15π12)−cos(19π12)]−i[sin(19π12)−sin(15π12)]
=[2sin((17pi)/12)sin((2pi)/12)]-i[2cos((17pi)/12)sin((2pi)/12)]=[2sin(17π12)sin(2π12)]−i[2cos(17π12)sin(2π12)]
=[2sin((17pi)/12)xx1/2]-i[2cos((17pi)/12)xx1/2]=[2sin(17π12)×12]−i[2cos(17π12)×12]
=sin((17pi)/12)-icos((17pi)/12)=sin(17π12)−icos(17π12)
Now
- sin((17pi)/12)=sqrt(1/2(1-cos((17pi)/6)))=sqrt(1/2(1-cos(3pi-pi/6))sin(17π12)=√12(1−cos(17π6))=√12(1−cos(3π−π6))
=sqrt(1/2(1+cos(pi/6)))=sqrt(1/2(1+cos(pi/6)))=√12(1+cos(π6))=√12(1+cos(π6))
=sqrt((2+sqrt3)/4)=sqrt((4+2sqrt3)/8)=(sqrt3+1)/(2sqrt2)=√2+√34=√4+2√38=√3+12√2
and
- cos((17pi)/12)=sqrt(1/2(1+cos((17pi)/6)))=sqrt(1/2(1+cos(3pi-pi/6))cos(17π12)=√12(1+cos(17π6))=√12(1+cos(3π−π6))
=sqrt(1/2(1-cos(pi/6)))=sqrt(1/2(1-cos(pi/6)))
=sqrt((2-sqrt3)/4)=sqrt((4-2sqrt3)/8)=(sqrt3-1)/(2sqrt2)
Hence
e^( ( 5 pi)/4 i) - e^( ( 19 pi)/12i)
=sin((17pi)/12)-icos((17pi)/12)
=1/(2sqrt2)((sqrt3+1)-i(sqrt3-1))
