How do you evaluate $e^{\frac{7 π}{4} i} - e^{\frac{4 π}{3} i}$ $e^( ( 7 pi)/4 i) - e^( ( 4 pi)/3 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{7 π}{4} i} - e^{\frac{4 π}{3} i}$ $e^( ( 7 pi)/4 i) - e^( ( 4 pi)/3 i)$ using trigonometric functions?

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Answer 1 · 2017-08-06T20:16:22

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

Explanation:

$e^{i x} = cos x + i sin x$ $e^(ix)=cosx+isinx$

Here, we have $e^{\frac{7 π}{4} i} - e^{\frac{4 π}{3} i} = cos (\frac{7 π}{4}) + i sin (\frac{7 π}{4}) - (cos (\frac{4 π}{3}) + i sin (\frac{4 π}{3})) = cos (\frac{7 π}{4}) + i sin (\frac{7 π}{4}) - cos (\frac{4 π}{3}) - i sin (\frac{4 π}{3})$ $e^((7pi)/4i)-e^((4pi)/3i)=cos((7pi)/4)+isin((7pi)/4)-(cos((4pi)/3)+isin((4pi)/3))=cos((7pi)/4)+isin((7pi)/4)-cos((4pi)/3)-isin((4pi)/3)$

$cos (\frac{7 π}{4}) = \frac{\sqrt{2}}{2}$ $cos((7pi)/4)=sqrt(2)/2$
$sin (\frac{7 π}{4}) = - \frac{\sqrt{2}}{2}$ $sin((7pi)/4)=-sqrt(2)/2$
$cos (\frac{4 π}{3}) = - \frac{1}{2}$ $cos((4pi)/3)=-1/2$
$sin (\frac{4 π}{3}) = - \frac{\sqrt{3}}{2}$ $sin((4pi)/3)=-sqrt(3)/2$

Therefore, $e^{\frac{7 π}{4} i} - e^{\frac{4 π}{3} i} = \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} + \frac{1}{2} + i \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{2}}{2} + i (\frac{- \sqrt{2} + \sqrt{3}}{2})$ $e^((7pi)/4i)-e^((4pi)/3i)=sqrt(2)/2-isqrt(2)/2+1/2+isqrt(3)/2=(1+sqrt(2))/2+i((-sqrt(2)+sqrt(3))/2)$

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How do you evaluate $e^{\frac{7 π}{4} i} - e^{\frac{4 π}{3} i}$ $e^( ( 7 pi)/4 i) - e^( ( 4 pi)/3 i)$ using trigonometric functions?

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