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How do you evaluate # e^( ( pi)/4 i) - e^( ( 7 pi)/6 i)# using trigonometric functions?

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Jan 9, 2018

#e^((pi i)/4)# = #cos (pi/4) + i sin (pi/4)= (sqrt2)/2 +i(sqrt2)/2#

#e^((7pi i)/6)#= # cos (7pi/6)+ isin (7pi/6)= -cos (pi/6)-isin (pi/6)#
#=-sqrt3/2 -i1/2#
(In the third quadrant both sin and cos are negative)

Thus, #e^((pi i)/4)# -#e^((7pi i)/6)#= #(sqrt2 +sqrt3)/2 +i(sqrt2 +1)/2#

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