How do you evaluate $e^{\frac{5 π}{4} i} - e^{\frac{11 π}{4} i}$ $e^( ( 5 pi)/4 i) - e^( ( 11 pi)/4 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{5 π}{4} i} - e^{\frac{11 π}{4} i}$ $e^( ( 5 pi)/4 i) - e^( ( 11 pi)/4 i)$ using trigonometric functions?

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Answer 1 · 2018-08-01T15:24:37

$= \sqrt{2} i$ $= sqrt2 i$

Explanation:

Use $e^{i π} = - 1 and e^{i 2 π} = 1$ $e^(i pi ) = - 1 and e^(i 2 pi) = 1$

$e^{i \frac{5}{4} π} - e^{i \frac{11}{4} π}$ $e^(i5/4pi) - e^(i11/4pi)$

$= e^{i \frac{π}{4}} e^{i π} - e^{- i \frac{π}{4}} e^{i 2 π}$ $= e^(ipi/4)e^(i pi) - e^(- i pi/4)e^(i 2 pi )$

$= e^{i \frac{π}{4}} (- 1) - e^{- i \frac{π}{4}} (1)$ $= e^(ipi/4)(-1) - e^(- i pi/4) ( 1 )$

$= (cos (\frac{π}{4}) + i s i m n (\frac{π}{4})) - (cos (\frac{π}{4}) - i sin (\frac{π}{4}))$ $= ( cos (pi/4) + i simn ( pi/4) ) -( cos (pi/4) - i sin ( pi/4))$

$= 2 sin (\frac{π}{4}) i$ $= 2 sin ( pi/4 ) i$

$\sqrt{2} i$ $sqrt2 i$

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How do you evaluate $e^{\frac{5 π}{4} i} - e^{\frac{11 π}{4} i}$ $e^( ( 5 pi)/4 i) - e^( ( 11 pi)/4 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{5 π}{4} i} - e^{\frac{11 π}{4} i}$ $e^( ( 5 pi)/4 i) - e^( ( 11 pi)/4 i)$ using trigonometric functions?