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

1 Answer

Jan 14, 2016

#-0.866-0.5i#

Explanation:

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#

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