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

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

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Answer 1 · 2016-11-09T17:55:59

$\frac{2}{\sqrt{2}}$ $2/sqrt2$

Explanation:

$e^{(\frac{π}{4}) i}$ $e^((pi/4)i$ = $cos (\frac{π}{4}) + i sin (\frac{π}{4}) = \frac{1 + i}{\sqrt{2}}$ $cos (pi/4) +i sin (pi/4)= (1+i)/sqrt2$

Like wise,

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

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

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

Hence $e^{(\frac{π}{4}) i}$ $e^((pi/4)i$ $- e^{\frac{3 π}{4} i}$ $-e^((3pi)/4i)$ = $\frac{1 + i}{\sqrt{2}} - \frac{- 1 + i}{\sqrt{2}}$ $(1+i)/sqrt2 - (-1+i)/sqrt 2$ = $\frac{2}{\sqrt{2}}$ $2/sqrt2$

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

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