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

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

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Answer 1 · 2018-05-10T13:47:46

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

Explanation:

A number given a $r e^{i θ}$ $re^(itheta)$ can be written as $r (cos θ + i sin θ)$ $r(costheta+isintheta)$

$r_{1} (cos (θ_{1}) + i sin (θ_{1})) \cdot r_{2} (cos (θ_{2}) + i sin (θ_{2})) = r_{1} r_{2} (cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2}))$ $r_1(cos(theta_1)+isin(theta_1))*r_2(cos(theta_2)+isin(theta_2))=r_1r_2(cos(theta_1+theta_2)+isin(theta_1+theta_2))$

$r_{1} r_{1} = 1 \cdot 1 = 1$ $r_1r_1=1*1=1$
$θ_{1} + θ_{2} = \frac{π}{3} + \frac{3 π}{2} = \frac{11 π}{6}$ $theta_1+theta_2=pi/3 +(3pi)/2=(11pi)/6$

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

$e^{\frac{11 π}{6} i}$ $e^((11pi)/6i$

Answer 2 · 2018-05-10T14:12:32

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

Explanation:

This is another point of view.

By Euler's relation

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

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

$e^{\frac{π}{3} i} = cos (\frac{π}{3}) + i sin (\frac{π}{3}) = \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $e^(pi/3i)=cos(pi/3)+isin(pi/3)=1/2+sqrt3/2i$

$e^{\frac{3}{2} π i} = cos (\frac{3}{2} π) + i sin (\frac{3}{2} π) = 0 - i$ $e^(3/2pii)=cos(3/2pi)+isin(3/2pi)=0-i$

Therefore,

$e^{\frac{π}{3} i} \cdot e^{\frac{3}{2} π i} = (\frac{1}{2} + \frac{\sqrt{3}}{2} i) \cdot (- I)$ $e^(pi/3i)*e^(3/2pii) = (1/2+sqrt3/2i)*(-I)$

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

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

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