How do you evaluate # e^( ( 7 pi)/4 i) - e^( ( 5 pi)/8 i)# using trigonometric functions?

1 Answer
Bdub
Mar 23, 2016

#e^((7pi)/4 i)-e^((5pi)/8 i) ~~1.0898-1.631i#

Explanation:

#e^(itheta) = cos theta +isin theta#

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

#e^((5pi)/8 i) = cos ((5pi)/8) +i sin ((5pi)/8)#

#e^((7pi)/4 i)-e^((5pi)/8 i)=[cos( (7pi)/4) +i sin( (7pi)/4)]-[cos ((5pi)/8) +i sin ((5pi)/8)]#

#e^((7pi)/4 i)-e^((5pi)/8 i) = [cos( (7pi)/4)-cos ((5pi)/8)]+i[sin( (7pi)/4)-sin ((5pi)/8)]#

#e^((7pi)/4 i)-e^((5pi)/8 i) ~~1.0898-1.631i#

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