How do you evaluate #e^( ( 13 pi)/8 i) - e^( ( pi)/2 i)# using trigonometric functions?

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Answer 1 · 2016-04-11T18:47:59

#e^((13pi)/8i) - e^(pi/2i) = 0.383 - 1.924i#

According to Euler's formula,

#e^(ix) = cosx + isinx#

Using values for #x# from the equation, #(13pi)/8# and #pi/2#

#x = (13pi)/8#
#e^((13pi)/8i) = cos((13pi)/8) + isin((13pi)/8)#
# = cos292.5 + isin292.5#
# = 0.383 - 0.924i#

#x = pi/2#
#e^(pi/2i) = cos(pi/2) + isin(pi/2)#
# = cos90 + isin90#
# = i#

Inserting these two values back into the original question,

#e^((13pi)/8i) - e^(pi/2i) = 0.383 - 0.924i - i#
# = 0.383 - 1.924i#