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.

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