How do you evaluate $e^{\frac{7 π}{6} i} - e^{\frac{π}{3} i}$ $e^( ( 7 pi)/6 i) - e^( ( pi)/3 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{7 π}{6} i} - e^{\frac{π}{3} i}$ $e^( ( 7 pi)/6 i) - e^( ( pi)/3 i)$ using trigonometric functions?

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Answer 1 · 2017-12-18T11:13:39

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

Explanation:

Using Euler's formula ( $e^{x i} = cos x - i sin x$ $e^(x i)=cosx-isinx$ ) we get:
$e^{\frac{7 π}{6} i} - e^{\frac{π}{3} i} = (cos (\frac{7 π}{6}) - i sin (\frac{7 π}{6})) - (cos (\frac{π}{3}) - i sin (\frac{π}{3}))$ $e^((7pi)/6i)-e^((pi)/3i)=(cos((7pi)/6)-isin((7pi)/6))-(cos(pi/3)-isin(pi/3))$

$cos (\frac{7 π}{6}) = - \frac{\sqrt{3}}{2}$ $cos((7pi)/6)=-sqrt(3)/2$
$sin (\frac{7 π}{6}) = - \frac{1}{2}$ $sin((7pi)/6)=-1/2$
$cos (\frac{π}{3}) = \frac{1}{2}$ $cos(pi/3)=1/2$
$sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $sin(pi/3)=sqrt(3)/2$

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

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

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

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

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How do you evaluate $e^{\frac{7 π}{6} i} - e^{\frac{π}{3} i}$ $e^( ( 7 pi)/6 i) - e^( ( pi)/3 i)$ using trigonometric functions?

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