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

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

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Answer 1 · 2016-03-27T08:35:51

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

Explanation:

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

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

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

${cos (\frac{3 π}{2}) + i sin (\frac{3 π}{2})} \cdot {cos (\frac{3 π}{2}) + i sin (\frac{3 π}{2})}$ ${cos((3pi)/2)+isin((3pi)/2)}*{cos((3pi)/2)+isin((3pi)/2)}$ or

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

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

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

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

= $cos 3 π + i sin 3 π$ $cos3pi+isin3pi$

= $- 1 + i \times 0 = - 1$ $-1+ixx0=-1$

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

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Explanation:

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