How do you evaluate $e^{\frac{15 π}{8} i} - e^{\frac{11 π}{6} i}$ $e^( ( 15 pi)/8 i) - e^( ( 11 pi)/6 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{15 π}{8} i} - e^{\frac{11 π}{6} i}$ $e^( ( 15 pi)/8 i) - e^( ( 11 pi)/6 i)$ using trigonometric functions?

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Answer 1 · 2016-03-28T09:30:25

$e^{\frac{15 π}{8} i} - e^{\frac{11 π}{6} i} = - 1.7899 + 0.8827 i$ $e^((15pi)/8i)-e^((11pi)/6i)=-1.7899+0.8827i$

Explanation:

As $e^{i θ} = cos θ + i sin θ$ $e^(itheta)=costheta+isintheta$ , we have

$e^{\frac{15 π}{8} i} = cos (\frac{15 π}{8}) + i sin (\frac{15 π}{8})$ $e^((15pi)/8i)=cos((15pi)/8)+isin((15pi)/8)$

= $cos (π - \frac{π}{8}) + i sin (π - \frac{π}{8})$ $cos(pi-pi/8)+isin(pi-pi/8)$

= $- cos (\frac{π}{8}) + i sin (\frac{π}{8}) = - 0.9239 + 0.3827 i$ $-cos(pi/8)+isin(pi/8)=-0.9239+0.3827i$

$e^{\frac{11 π}{6} i} = cos (\frac{11 π}{6}) + i sin (\frac{11 π}{6})$ $e^((11pi)/6i)=cos((11pi)/6)+isin((11pi)/6)$

= $cos (2 π - \frac{π}{6}) + i sin (2 π - \frac{π}{6})$ $cos(2pi-pi/6)+isin(2pi-pi/6)$

= $cos (\frac{π}{6}) - i sin (\frac{π}{6})$ $cos(pi/6)-isin(pi/6)$

= $\frac{\sqrt{3}}{2} - i \cdot \frac{1}{2} = 0.8660 - 0.5 i$ $sqrt3/2-i*1/2=0.8660-0.5i$

Hence $e^{\frac{15 π}{8} i} - e^{\frac{11 π}{6} i} = (- 0.9239 + 0.3827 i) - (0.8660 - 0.5 i)$ $e^((15pi)/8i)-e^((11pi)/6i)=(-0.9239+0.3827i)-(0.8660-0.5i)$

= $(- 0.9239 - 0.8660) + i (0.3827 + 0.5)$ $(-0.9239-0.8660)+i(0.3827+0.5)$

= $- 1.7899 + 0.8827 i$ $-1.7899+0.8827i$

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How do you evaluate $e^{\frac{15 π}{8} i} - e^{\frac{11 π}{6} i}$ $e^( ( 15 pi)/8 i) - e^( ( 11 pi)/6 i)$ using trigonometric functions?

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Explanation:

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How do you evaluate $e^{\frac{15 π}{8} i} - e^{\frac{11 π}{6} i}$ $e^( ( 15 pi)/8 i) - e^( ( 11 pi)/6 i)$ using trigonometric functions?