How do you evaluate $e^( (3 pi)/2 i) - e^( ( 11 pi)/6 i)$ using trigonometric functions?

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Answer 1 · 2016-01-14T21:21:54

$-0.866-0.5i$

Euler's Theorem states that

$e^(itheta)=costheta+isintheta$ .

Application hereof to the given example yields

$e^((3pi)/2i)-e^((11pi)/6i)=cos((3pi)/2)+isin((3pi)/2)-cos((11pi)/6)-isin((11pi)/6)$

$=0-i-0.866+0.5i$

$=-0.866-0.5i$

How do you evaluate e^( (3 pi)/2 i) - e^( ( 11 pi)/6 i) e^( (3 pi)/2 i) - e^( ( 11 pi)/6 i) using trigonometric functions?