How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{11 π}{6} i}$ $e^( ( pi)/4 i) - e^( ( 11 pi)/6 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{11 π}{6} i}$ $e^( ( pi)/4 i) - e^( ( 11 pi)/6 i)$ using trigonometric functions?

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Answer 1 · 2017-05-26T02:16:45

$e^{\frac{π}{4} i} - e^{\frac{11 π}{6} i} = \frac{\sqrt{2} - \sqrt{3}}{2} + \frac{\sqrt{2} + 1}{2} i$ $e^(pi/4i)-e^((11pi)/6i)=(sqrt2-sqrt3)/2 + (sqrt2+1)/2i$

Explanation:

There is a special formula for complex numbers that we will use to solve this problem:

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

Therefore, we can rewrite this problem as:

$e^{\frac{π}{4} i} - e^{\frac{11 π}{6} i}$ $e^(pi/4i)-e^((11pi)/6i)$

$= cos (\frac{π}{4}) + i sin (\frac{π}{4}) - cos (\frac{11 π}{6}) - i sin (\frac{11 π}{6})$ $=cos(pi/4)+isin(pi/4)-cos((11pi)/6)-isin((11pi)/6)$

Now all we have to do is simplify.

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

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

So $e^{\frac{π}{4} i} - e^{\frac{11 π}{6} i} = \frac{\sqrt{2} - \sqrt{3}}{2} + \frac{\sqrt{2} + 1}{2} i$ $e^(pi/4i)-e^((11pi)/6i)=(sqrt2-sqrt3)/2 + (sqrt2+1)/2i$

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{11 π}{6} i}$ $e^( ( pi)/4 i) - e^( ( 11 pi)/6 i)$ using trigonometric functions?

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Explanation:

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