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

1 Answer
Narad T.
Oct 17, 2016

#e^((5ipi)/4)-e^((7ipi)/4)=-sqrt2#

Explanation:

Use the definition #e^(itheta)=costheta+isintheta#
#e^((5ipi)/4)=cos((5pi)/4)+isin((5pi)/4)#
#e^((7ipi)/4)=cos((7pi)/4)+isin((7pi)/4)#
#cos((5pi)/4)=-sqrt2/2#
#cos((7pi)/4)=sqrt2/2#
#sin((7pi)/4)=-sqrt2/2#
#sin((5pi)/4)=-sqrt2/2#
The result is
#e^((5ipi)/4)-e^((7ipi)/4)=cos((5pi)/4)+isin((5pi)/4)-(cos((7pi)/4)+isin((7pi)/4))#
#e^((5ipi)/4)-e^((7ipi)/4)=-sqrt2/2-isqrt2/2-sqrt2/2+isqrt2/2#
So the final answer is
#e^((5ipi)/4)-e^((7ipi)/4)=-sqrt2#

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