How do you evaluate # e^( ( 13 pi)/8 i) - e^( (5 pi)/4 i)# using trigonometric functions?

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Answer 1 · 2018-07-27T14:01:18

#color(crimson)(e^(((13pi)/8)i) - e^(((5pi)/4)i) = -0.3244 - 1.631#, III 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^(((5pi)/4)i) = cos ((5pi)/4) + i sin ((5pi)/4)#

#=> -0.7071 - 0.7071 i#, III Quadrant.

#e^(((13pi)/8)i) - e^(((5pi)/4)i) = 0.3827 - 0.7071 - 0.9239i - 0.7071i#

#=> -0.3244 - 1.631#, III Quadrant.