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

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Answer 1 · 2016-04-09T17:32:03

#1.006 + 0.824i#

According to Euler's formula,

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

Using values for #x# from the equation,

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

#x = (5pi)/3#
#e^((5pi)/3i) = cos((5pi)/3) + isin((5pi)/3)#
# = cos300 + isin300#
# = - 0.022 - i#

Inserting these two values into the equation above,

#e^((3pi)/2i) - e^((5pi)/3i) = 0.984 - 0.176i + 0.022 + i#
# = 1.006 + 0.824i#