How do you evaluate $e^{\frac{3 π}{2} i} - e^{\frac{13 π}{8} i}$ $e^( (3 pi)/2 i) - e^( (13 pi)/8 i)$ using trigonometric functions?

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Answer 1 · 2018-07-27T14:20:45

$\Rightarrow - 0.3827 - 0.0761 i$ $color(chocolate)(=> -0.3827 - 0.0761 i$ , IV Quadrant.

$e^{i θ} = cos θ + i sin θ$ $e^(i theta) = cos theta + i sin theta$

$e^{(\frac{3 π}{2}) i} = cos (\frac{3 π}{2}) + i sin (\frac{3 π}{2})$ $e^(((3pi)/2) i )= cos ((3pi)/2) + i sin ((3pi)/2)$

$\approx > - i$ $~~> - i$ , III Quadrant

$e^{(\frac{13 π}{8}) i} = cos (\frac{13 π}{8}) + i sin (\frac{13 π}{8})$ $e^(((13pi)/8)i) = cos ((13pi)/8) + i sin ((13pi)/8)$

$\Rightarrow 0.3827 - 0.9239 i$ $=> 0.3827 - 0.9239 i$ , IV Quadrant.

$e^{(\frac{3 π}{2}) i} - e^{(\frac{13 π}{8}) i} = - 0.3827 - i + 0.9239 i$ $e^(((3pi)/2)i) - e^(((13pi)/8)i) = -0.3827 - i + 0.9239 i$

$\Rightarrow - 0.3827 - 0.0761 i$ $color(chocolate)(=> -0.3827 - 0.0761 i$ , IV Quadrant.

How do you evaluate e3π2i−e13π8i e^( (3 pi)/2 i) - e^( (13 pi)/8 i) using trigonometric functions?