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

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

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Answer 1 · 2016-03-09T03:52:17

$e^{i \frac{3 π}{4}} \cdot e^{i \frac{3 π}{2}} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$ $e^{i{3pi}/4} * e^{i{3pi}/2} = 1/sqrt2+ i/sqrt2$

Explanation:

From the identity

$e^{i θ} \equiv cos (θ) + i sin (θ)$ $e^{itheta} -= cos(theta) + i sin(theta)$

We write

$e^{i \frac{3 π}{4}} \cdot e^{i \frac{3 π}{2}} = (cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4})) \cdot (cos (\frac{3 π}{2}) + i sin (\frac{3 π}{2}))$ $e^{i{3pi}/4} * e^{i{3pi}/2} = (cos({3pi}/4) + i sin({3pi}/4)) * (cos({3pi}/2) + i sin({3pi}/2))$

$= (- \frac{1}{\sqrt{2}} + i (\frac{1}{\sqrt{2}})) \cdot (0 + i (- 1))$ $= (-1/sqrt2 + i (1/sqrt2)) * (0 + i (-1))$

$= \frac{1}{\sqrt{2}} \cdot (- 1 + i) \cdot (- i)$ $= 1/sqrt2*(-1 + i) * (- i)$

$= \frac{1}{\sqrt{2}} \cdot (i + 1)$ $= 1/sqrt2*(i + 1)$

$= \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$ $= 1/sqrt2+ i/sqrt2$

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

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Explanation:

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