How do you evaluate # e^( ( 5 pi)/4 i) - e^( ( 3 pi)/4 i)# using trigonometric functions?

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Answer 1 · 2018-01-28T17:17:53

#sqrt(2)e^((3pi)/2*i)#

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

#e^((5pi)/4i) = cos((5pi)/4)+isin((5pi)/4) = -sqrt(2)/2-sqrt(2)/2i#
and
#e^((3pi)/4i) = cos((3pi)/4)+isin((3pi)/4) = -sqrt(2)/2+sqrt(2)/2i#

Which means that

#e^((5pi)/4i) - e^((3pi)/4i) = -sqrt(2)/2-sqrt(2)/2i - (-sqrt(2)/2+sqrt(2)/2i)#

#=-sqrt(2)/2-sqrt(2)/2i +sqrt(2)/2-sqrt(2)/2i#

#=-sqrt(2)i#

which is a complex number with #r = sqrt(2)# and #theta = (3pi)/2#

so

#e^((5pi)/4i) - e^((3pi)/4i) = sqrt(2)e^((3pi)/2*i)#