How do you evaluate e^( ( 7 pi)/4 i) - e^( ( 5 pi)/8 i)e7π4i−e5π8i using trigonometric functions?

1 Answer
Bdub
Mar 23, 2016

e^((7pi)/4 i)-e^((5pi)/8 i) ~~1.0898-1.631ie7π4i−e5π8i≈1.0898−1.631i

Explanation:

e^(itheta) = cos theta +isin thetaeiθ=cosθ+isinθ

e^((7pi)/4 i) = cos( (7pi)/4) +i sin( (7pi)/4)e7π4i=cos(7π4)+isin(7π4)

e^((5pi)/8 i) = cos ((5pi)/8) +i sin ((5pi)/8)e5π8i=cos(5π8)+isin(5π8)

e^((7pi)/4 i)-e^((5pi)/8 i)=[cos( (7pi)/4) +i sin( (7pi)/4)]-[cos ((5pi)/8) +i sin ((5pi)/8)]e7π4i−e5π8i=[cos(7π4)+isin(7π4)]−[cos(5π8)+isin(5π8)]

e^((7pi)/4 i)-e^((5pi)/8 i) = [cos( (7pi)/4)-cos ((5pi)/8)]+i[sin( (7pi)/4)-sin ((5pi)/8)]e7π4i−e5π8i=[cos(7π4)−cos(5π8)]+i[sin(7π4)−sin(5π8)]

e^((7pi)/4 i)-e^((5pi)/8 i) ~~1.0898-1.631ie7π4i−e5π8i≈1.0898−1.631i

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