How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{5 π}{3} i}$ $e^( ( pi)/4 i) - e^( ( 5 pi)/3 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{5 π}{3} i}$ $e^( ( pi)/4 i) - e^( ( 5 pi)/3 i)$ using trigonometric functions?

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Answer 1 · 2017-10-16T07:46:33

$e^{\frac{π}{4} i} - e^{\frac{5 π}{3} i} = \frac{\sqrt{2} - 1}{2} + i \frac{3 \sqrt{2} + 2 \sqrt{3}}{6}$ $e^( ( pi)/4 i) - e^( ( 5 pi)/3 i)=(sqrt(2)-1)/2+i(3sqrt(2)+2sqrt(3))/6$

Explanation:

Using Euler's identity:
$e^{i x} = cos x + i sin x$ $e^(ix)=cosx+isinx$

So, $e^{\frac{π}{4} i} - e^{\frac{5 π}{3} i} = (cos (\frac{π}{4}) + i sin (\frac{π}{4})) - (cos (\frac{5 π}{3}) + i sin (\frac{5 π}{3})$ $e^( ( pi)/4 i) - e^( ( 5 pi)/3 i)=(cos(\pi/4)+isin(\pi/4))-(cos((5\pi)/3)+isin((5\pi)/3)$
$= (\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}) - (\frac{1}{2} - i \frac{\sqrt{3}}{3})$ $=(sqrt(2)/2+isqrt(2)/2)-(1/2-isqrt(3)/3)$
$= \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} - \frac{1}{2} + i \frac{\sqrt{3}}{3}$ $=sqrt(2)/2+isqrt(2)/2-1/2+isqrt(3)/3$
$= \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} - \frac{1}{2} + i \frac{\sqrt{3}}{3}$ $=sqrt(2)/2+isqrt(2)/2-1/2+isqrt(3)/3$
$= \frac{\sqrt{2} - 1}{2} + i \frac{3 \sqrt{2} + 2 \sqrt{3}}{6}$ $=(sqrt(2)-1)/2+i(3sqrt(2)+2sqrt(3))/6$

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{5 π}{3} i}$ $e^( ( pi)/4 i) - e^( ( 5 pi)/3 i)$ using trigonometric functions?

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