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

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Answer 1 · 2018-08-01T15:37:11

#color(chocolate)(=> 0.3827 + 0.0761 i #, I Quadrant.

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

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

#~~> 0.3827 - 0.9239 i #, IV Quadrant

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

#=> - i#, III Quadrant.

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

#color(chocolate)(=> 0.3827 + 0.0761 i #, I Quadrant.