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

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

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Answer 1 · 2016-11-03T13:00:41

$\sqrt[2]{2} + i \sqrt[2]{2} = 2 (cos (\frac{π}{4}) + i sin (\frac{π}{4}) = 2 e^{i \frac{π}{4}}$ $root2(2)+iroot2(2)=2(cos(pi/4)+isin(pi/4)=2e^(ipi/4)$

Explanation:

$cos (\frac{π}{4}) + i sin (\frac{π}{4}) - cos (5 \frac{π}{4}) - i sin (5 \frac{π}{4})$ $Cos(pi/4)+isin(pi/4)-cos(5pi/4)-isin(5pi/4)$

$\frac{\sqrt[2]{2}}{2} + i \frac{\sqrt[2]{2}}{2} - (- \frac{\sqrt[2]{2}}{2}) - i (- \frac{\sqrt[2]{2}}{2}) = \sqrt[2]{2} + i \sqrt[2]{2}$ $root2(2)/2+iroot2(2)/2-(-root2(2)/2)-i(-root2(2)/2)=root2(2)+iroot2(2)$

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

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