How do you evaluate # e^( ( 13 pi)/8 i) - e^( (17 pi)/12 i)# using trigonometric functions?

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Answer 1 · 2016-03-26T05:01:48

#e^((13pi)/8i)-e^((17pi)/12i)=0.0003+0.01142i##

As #e^(itheta)=costheta+isintheta#

#e^((13pi)/8i)=cos((13pi)/8)+isin((13pi)/8)#

= #cos((13pi)/8-2pi)+isin((13pi)/8-2pi)#

= #cos((-3pi)/8)+isin((-3pi)/8)#

= #0.99979-0.02056i# (using scientific calculator)

Similarly #e^((17pi)/12i)=cos((17pi)/12)+isin((17pi)/12)#

= #cos((17pi)/12-2pi)+isin((17pi)/12-2pi)#

= #cos((-7pi)/12)+isin((-7pi)/12)#

= #0.99949-0.03198i# (using scientific calculator)

Hence #e^((13pi)/8i)-e^((17pi)/12i)#

= #(0.99979-0.02056i)-(0.99949-0.03198i)#

= #(0.99979-0.99949)-(0.02056-0.03198)i#

= #0.0003-(-0.01142)i#

= #0.0003+0.01142i#