How do you multiply $e^{\frac{4 π}{3} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 4 pi )/ 3 i) * e^( 3 pi/2 i )$ in trigonometric form?

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Answer 1 · 2018-04-22T13:19:47

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

Apply Euler's identity

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

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

$cos (\frac{4}{3} π) = - \frac{1}{2}$ $cos(4/3pi)=-1/2$

$sin (\frac{4}{3} π) = - \frac{\sqrt{3}}{2}$ $sin(4/3pi)=-sqrt3/2$

$sin (\frac{3}{2} π) = - 1$ $sin(3/2pi)=-1$

$cos (\frac{3}{2} π) = 0$ $cos(3/2pi)=0$

Therefore,

$e^{\frac{4}{3} i π} . e^{\frac{3}{2} i π}$ $e^(4/3ipi).e^(3/2ipi)$

$= (cos (\frac{4}{3} π) + i sin (\frac{4}{3} π)) (cos (\frac{3}{2} π) + i sin (\frac{3}{2} π))$ $=(cos(4/3pi)+isin(4/3pi))(cos(3/2pi)+isin(3/2pi))$

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

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

How do you multiply e4π3i⋅e3π2ie^(( 4 pi )/ 3 i) * e^( 3 pi/2 i ) in trigonometric form?