How do you evaluate # e^( (3 pi)/2 i) - e^( (13 pi)/8 i)# using trigonometric functions?

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Answer 1 · 2018-07-27T14:20:45

#color(chocolate)(=> -0.3827 - 0.0761 i #, IV Quadrant.

#e^(i theta) = cos theta + i sin theta#

#e^(((3pi)/2) i )= cos ((3pi)/2) + i sin ((3pi)/2)#

#~~> - i#, III Quadrant

#e^(((13pi)/8)i) = cos ((13pi)/8) + i sin ((13pi)/8)#

#=> 0.3827 - 0.9239 i#, IV Quadrant.

#e^(((3pi)/2)i) - e^(((13pi)/8)i) = -0.3827 - i + 0.9239 i #

#color(chocolate)(=> -0.3827 - 0.0761 i #, IV Quadrant.