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.

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