How do you evaluate # e^( ( 13 pi)/8 i) - e^( ( 5 pi)/3 i)# using trigonometric functions?

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Answer 1 · 2018-07-27T14:30:21

#color(maroon)(=> -0.1173 - 0.0579 i #, III Quadrant.

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

#e^(((5pi)/3) i )= cos ((5pi)/3) + i sin ((5pi)/3)#

#~~> 0.5 - 0.866 i#, IV Quadrant

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

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

#e^(((pi)/12)i) - e^(((pi)/8)i) = -0.3827 - 0.9239 i - 0.5 + 0.866 i #

#color(chocolate)(=> -0.1173 - 0.0579 i #, III Quadrant.