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

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

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Answer 1 · 2016-05-19T10:30:33

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

Explanation:

$a + i b$ $a+ib$ can be written in trigonometric form $r e^{i θ} = r cos θ + i r sin θ = r (cos θ + i sin θ)$ $re^(itheta)=rcostheta+irsintheta=r(costheta+isintheta)$ ,
where $r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$ , but here in given question $r = 1$ $r=1$

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

= $e^{(\frac{4 π}{6} + \frac{9 π}{6}) i} = e^{13 \frac{π}{6} i}$ $e^(((4pi)/6+(9pi)/6)i)=e^(13pi/6i)$

= $(cos (13 \frac{π}{6}) + i sin (13 \frac{π}{6}))$ $(cos(13pi/6)+isin(13pi/6))$

= $(cos (2 π + \frac{π}{6}) + i sin (2 π + \frac{π}{6}))$ $(cos(2pi+pi/6)+isin(2pi+pi/6))$

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

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

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

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Explanation:

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