How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{2 π}{3} i}$ $e^( ( pi)/4 i) - e^( ( 2 pi)/3 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{2 π}{3} i}$ $e^( ( pi)/4 i) - e^( ( 2 pi)/3 i)$ using trigonometric functions?

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Answer 1 · 2016-07-02T20:24:56

Recall Euler's formula:

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

Using this:

$e^{\frac{i π}{4}} - e^{\frac{2 i π}{3}} = cos (\frac{π}{4}) + i sin (\frac{π}{4}) - cos (\frac{2 π}{3}) - i sin (\frac{2 π}{3})$ $e^((ipi)/4) - e^((2ipi)/3) = cos(pi/4) + isin(pi/4) - cos((2pi)/3) - isin((2pi)/3)$

Evaluating these trig functions we get:

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

And grouping terms together gives us

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

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{2 π}{3} i}$ $e^( ( pi)/4 i) - e^( ( 2 pi)/3 i)$ using trigonometric functions?

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