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

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

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Answer 1 · 2017-02-26T06:27:14

Use Euler's identity:

$e^{i x} = cos x + i sin x$ $e^(ix) = cosx + isinx$

Thus:

$e^{\frac{3 π}{2} i} - e^{\frac{π}{3} i}$ $e^((3pi)/2i) - e^(pi/3i)$

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

$= cancel(cos((3pi)/2))^(0) + isin((3pi)/2) - cos(pi/3) - isin(pi/3)$

$= -i - 1/2 - sqrt3/2i$

$= color(blue)(-1/2 + (-1 - sqrt3/2)i)$

So, for the form $a + bi$ , $a = -1/2$ and $b = -1 - sqrt3/2$ .

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

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