How do you multiply $e^{\frac{11 π}{12}} \cdot e^{\frac{π}{4} i}$ $e^((11pi )/ 12 ) * e^( pi/4 i )$ in trigonometric form?

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How do you multiply $e^{\frac{11 π}{12}} \cdot e^{\frac{π}{4} i}$ $e^((11pi )/ 12 ) * e^( pi/4 i )$ in trigonometric form?

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Answer 1 · 2018-05-30T08:23:43

The answer is $= - \frac{\sqrt{3}}{2} - \frac{1}{2} i$ $=-sqrt3/2-1/2i$

Explanation:

Apply Euler's relation

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

$e^{\frac{11}{12} i π} = cos (\frac{11}{12} π) + i sin (\frac{11}{12} π)$ $e^(11/12ipi)=cos(11/12pi)+isin(11/12pi)$

$e^{\frac{1}{4} i π} = cos (\frac{1}{4} π) + i sin (\frac{1}{4} π)$ $e^(1/4ipi)=cos(1/4pi)+isin(1/4pi)$

$i^{2} = - 1$ $i^2=-1$

Therefore,

$e^{\frac{11}{12} i π} \cdot e^{\frac{1}{4} i π} = (cos (\frac{11}{12} π) + i sin (\frac{11}{12} π)) \cdot (cos (\frac{1}{4} π) + i sin (\frac{1}{4} π))$ $e^(11/12ipi)*e^(1/4ipi)=(cos(11/12pi)+isin(11/12pi))*(cos(1/4pi)+isin(1/4pi))$

$= cos (\frac{11}{12} π) cos (\frac{1}{4} π) - sin (\frac{11}{12} π) sin (\frac{1}{4} π) + i (cos (\frac{11}{12} π) sin (\frac{1}{4} π) + sin (\frac{11}{12} π) cos (\frac{1}{4} π)))$ $=cos(11/12pi)cos(1/4pi)-sin(11/12pi)sin(1/4pi)+i(cos(11/12pi)sin(1/4pi)+sin(11/12pi)cos(1/4pi)))$

$= cos (\frac{11}{12} π + \frac{1}{4} π) + i sin (\frac{11}{12} π + \frac{1}{4} π)$ $=cos(11/12pi+1/4pi)+isin(11/12pi+1/4pi)$

$= cos (\frac{14}{12} π) + i sin (\frac{14}{12} π)$ $=cos(14/12pi)+isin(14/12pi)$

$= cos (\frac{7}{6} π) + i sin (\frac{7}{6} π)$ $=cos(7/6pi)+isin(7/6pi)$

$= - \frac{\sqrt{3}}{2} - \frac{1}{2} i$ $=-sqrt3/2-1/2i$

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How do you multiply $e^{\frac{11 π}{12}} \cdot e^{\frac{π}{4} i}$ $e^((11pi )/ 12 ) * e^( pi/4 i )$ in trigonometric form?

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How do you multiply e^((11pi )/ 12 ) * e^( pi/4 i ) e11π12⋅eπ4ie^((11pi )/ 12 ) * e^( pi/4 i ) in trigonometric form?

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How do you multiply $e^{\frac{11 π}{12}} \cdot e^{\frac{π}{4} i}$ $e^((11pi )/ 12 ) * e^( pi/4 i )$ in trigonometric form?