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

1 Answer
Mar 26, 2016

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

Explanation:

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