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

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

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

= $0.0003-(-0.01142)i$

= $0.0003+0.01142i$

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