How do you multiply $e^{\frac{9 π}{4} i} \cdot e^{\frac{π}{2} i}$ $e^(( 9 pi )/ 4 i) * e^( pi/2 i )$ in trigonometric form?

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How do you multiply $e^{\frac{9 π}{4} i} \cdot e^{\frac{π}{2} i}$ $e^(( 9 pi )/ 4 i) * e^( pi/2 i )$ in trigonometric form?

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Answer 1 · 2016-09-26T10:23:14

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

$e^{\frac{9 π}{4} i} = cos (\frac{9 π}{4}) + i sin (\frac{9 π}{4})$ $e^((9pi)/4i)=cos((9pi)/4)+isin((9pi)/4)$ and

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

Hence $e^{\frac{9 π}{4} i} \cdot e^{\frac{π}{2} i} = (cos (\frac{9 π}{4}) + i sin (\frac{9 π}{4})) (cos (\frac{π}{2}) + i sin (\frac{π}{2}))$ $e^((9pi)/4i)*e^(pi/2i)=(cos((9pi)/4)+isin((9pi)/4))(cos(pi/2)+isin(pi/2))$

= $cos (\frac{9 π}{4}) (cos (\frac{π}{2}) + i sin (\frac{π}{2})) + i sin (\frac{9 π}{4})) (cos (\frac{π}{2}) + i sin (\frac{π}{2}))$ $cos((9pi)/4)(cos(pi/2)+isin(pi/2))+isin((9pi)/4))(cos(pi/2)+isin(pi/2))$

= $cos (\frac{9 π}{4}) cos (\frac{π}{2}) + i cos (\frac{9 π}{4}) sin (\frac{π}{2})) + i sin (\frac{9 π}{4}) cos (\frac{π}{2}) + i^{2} sin (\frac{9 π}{4}) sin (\frac{π}{2}))$ $cos((9pi)/4)cos(pi/2)+icos((9pi)/4)sin(pi/2))+isin((9pi)/4)cos(pi/2)+i^2sin((9pi)/4)sin(pi/2))$

= $cos (\frac{9 π}{4}) cos (\frac{π}{2}) + i cos (\frac{9 π}{4}) sin (\frac{π}{2})) + i sin (\frac{9 π}{4}) cos (\frac{π}{2}) - sin (\frac{9 π}{4}) sin (\frac{π}{2}))$ $cos((9pi)/4)cos(pi/2)+icos((9pi)/4)sin(pi/2))+isin((9pi)/4)cos(pi/2)-sin((9pi)/4)sin(pi/2))$

= $(cos (\frac{9 π}{4}) cos (\frac{π}{2}) - sin (\frac{9 π}{4}) sin (\frac{π}{2})) + i (sin (\frac{9 π}{4}) cos (\frac{π}{2}) + cos (\frac{9 π}{4}) sin (\frac{π}{2}))$ $(cos((9pi)/4)cos(pi/2)-sin((9pi)/4)sin(pi/2))+i(sin((9pi)/4)cos(pi/2)+cos((9pi)/4)sin(pi/2))$

= $cos ((\frac{9 π}{4}) + (\frac{π}{2})) + i sin ((\frac{9 π}{4}) + (\frac{π}{2}))$ $cos(((9pi)/4)+(pi/2))+isin(((9pi)/4)+(pi/2))$

= $cos (\frac{11 π}{4}) + i sin (\frac{11 π}{4})$ $cos((11pi)/4)+isin((11pi)/4)$

= $e^{\frac{11 π}{4}}$ $e^((11pi)/4)$

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How do you multiply $e^{\frac{9 π}{4} i} \cdot e^{\frac{π}{2} i}$ $e^(( 9 pi )/ 4 i) * e^( pi/2 i )$ in trigonometric form?

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