How do you multiply $e^{\frac{2 π}{3} i} \cdot e^{\frac{π}{2} i}$ $e^(( 2 pi )/ 3 i) * e^( pi/2 i )$ in trigonometric form?

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How do you multiply $e^{\frac{2 π}{3} i} \cdot e^{\frac{π}{2} i}$ $e^(( 2 pi )/ 3 i) * e^( pi/2 i )$ in trigonometric form?

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Answer 1 · 2018-05-15T13:36:35

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

Explanation:

$e^{i θ} = cos (θ) + i sin (θ)$ $e^(itheta)=cos(theta)+isin(theta)$

$e_{1}^{i θ_{1}} \cdot e_{2}^{i θ_{2}} = = cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2})$ $e^(itheta_1)*e^(itheta_2)==cos(theta_1+theta_2)+isin(theta_1+theta_2)$

$θ_{1} + θ_{2} = \frac{2 π}{3} + \frac{π}{2} = \frac{7 π}{6}$ $theta_1+theta_2=(2pi)/3+pi/2=(7pi)/6$

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

Answer 2 · 2018-05-15T13:56:16

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

Explanation:

Another method .

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

Euler's relation

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

Therefore,

$e^{\frac{2}{3} π i} \cdot e^{\frac{π}{2} i} = (cos (\frac{2}{3} π) + i sin (\frac{2}{3} π)) (cos (\frac{π}{2}) + i sin (\frac{π}{2}))$ $e^(2/3pii)*e^(pi/2i)=(cos(2/3pi)+isin(2/3pi))(cos(pi/2)+isin(pi/2))$

$= (\frac{1}{2} + i \frac{\sqrt{3}}{2}) (0 + i)$ $=(1/2+isqrt3/2)(0+i)$

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

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

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How do you multiply $e^{\frac{2 π}{3} i} \cdot e^{\frac{π}{2} i}$ $e^(( 2 pi )/ 3 i) * e^( pi/2 i )$ in trigonometric form?

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