How do you evaluate # e^( (3 pi)/2 i) - e^( ( 5 pi)/8 i)# using trigonometric functions?

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Answer 1 · 2016-04-08T22:22:18

#0.157 + 0.386i#

According to Euler's formula,

#e^(ix) = cosx + isinx#.

Inserting values for #x = (3pi)/2# and #(5pi)/8#,

#e^((3pi)/2i) = cos((3pi)/2) + isin((3pi)/2)#
# = cos270 + isin270#
# = 0.984 - 0.176i#

#e^((5pi)/8i) = cos((5pi)/8) + isin((5pi)/8)#
# = cos112.5 + isin112.5#
# = 0.827 - 0.562i#

Putting both of these together,

#e^((3pi)/2i) - e^((5pi)/8i) = 0.984 - 0.827 - 0.176i + 0.562i#
# = 0.157 + 0.386i#