How do you evaluate # e^( ( 7 pi)/4 i) - e^( ( 11 pi)/6 i)# using trigonometric functions?

1 Answer
Bdub
Mar 23, 2016

#e^((7pi)/4 i) -e^((11pi)/6 i) = -0.159-0.207i#

Explanation:

#e^(itheta) = cos theta +i sin theta#

#e^((7pi)/4 i) = cos ((7pi)/4)+isin((7pi)/4)#

#e^((11pi)/6 i) = cos ((11pi)/6)+isin((11pi)/6)#

#e^((7pi)/4 i) -e^((11pi)/6 i) =[cos ((7pi)/4)+isin((7pi)/4)]-[cos ((11pi)/6)+isin((11pi)/6)]#

#e^((7pi)/4 i) -e^((11pi)/6 i) =[cos ((7pi)/4)-cos ((11pi)/6)]+i[sin((7pi)/4)-sin((11pi)/6)]#

#e^((7pi)/4 i) -e^((11pi)/6 i) = -0.159-0.207i#

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