How do you evaluate # e^( ( 15 pi)/8 i) - e^( ( 11 pi)/6 i)# using trigonometric functions?

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Answer 1 · 2016-03-28T09:30:25

#e^((15pi)/8i)-e^((11pi)/6i)=-1.7899+0.8827i#

As #e^(itheta)=costheta+isintheta#, we have

#e^((15pi)/8i)=cos((15pi)/8)+isin((15pi)/8)#

= #cos(pi-pi/8)+isin(pi-pi/8)#

= #-cos(pi/8)+isin(pi/8)=-0.9239+0.3827i#

#e^((11pi)/6i)=cos((11pi)/6)+isin((11pi)/6)#

= #cos(2pi-pi/6)+isin(2pi-pi/6)#

= #cos(pi/6)-isin(pi/6)#

= #sqrt3/2-i*1/2=0.8660-0.5i#

Hence #e^((15pi)/8i)-e^((11pi)/6i)=(-0.9239+0.3827i)-(0.8660-0.5i)#

= #(-0.9239-0.8660)+i(0.3827+0.5)#

= #-1.7899+0.8827i#