How do you evaluate $e^{\frac{7 π}{4} i} - e^{\frac{π}{6} i}$ $e^( ( 7 pi)/4 i) - e^( ( pi)/6 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{7 π}{4} i} - e^{\frac{π}{6} i}$ $e^( ( 7 pi)/4 i) - e^( ( pi)/6 i)$ using trigonometric functions?

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Answer 1 · 2016-01-31T03:28:28

$e^{i \frac{7 π}{4}} - e^{i \frac{π}{6}} = \frac{\sqrt{2} - \sqrt{3}}{2} - i (\frac{1 + \sqrt{2}}{2})$ $e^{i (7pi)/4} - e^{i pi/6} = (sqrt2-sqrt3)/2 - i ((1+sqrt2)/2)$

Explanation:

Use the Euler's Formula, which states that $e^{i θ} \equiv cos (θ) + i sin (θ)$ $e^{i theta} -= cos(theta) + i sin(theta)$ . (Proof omitted)

Therefore,

$e^{i \frac{7 π}{4}} - e^{i \frac{π}{6}} = (cos (\frac{7 π}{4}) + i sin (\frac{7 π}{4})) - (cos (\frac{π}{6}) + i sin (\frac{π}{6}))$ $e^{i (7pi)/4} - e^{i pi/6} = (cos((7pi)/4) + i sin((7pi)/4)) - (cos(pi/6) + i sin(pi/6))$

$= (\frac{\sqrt{2}}{2} + i (- \frac{\sqrt{2}}{2})) - (\frac{\sqrt{3}}{2} + i (\frac{1}{2}))$ $= (sqrt2/2 + i (-sqrt2/2)) - (sqrt3/2 + i (1/2))$

$= \frac{\sqrt{2} - \sqrt{3}}{2} - i (\frac{1 + \sqrt{2}}{2})$ $= (sqrt2-sqrt3)/2 - i ((1+sqrt2)/2)$

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How do you evaluate $e^{\frac{7 π}{4} i} - e^{\frac{π}{6} i}$ $e^( ( 7 pi)/4 i) - e^( ( pi)/6 i)$ using trigonometric functions?

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