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

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

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Answer 1 · 2016-11-04T06:19:17

$e^{\frac{π}{4} i} \cdot e^{\frac{π}{2} i} = - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i$ $e^(pi/4i)*e^(pi/2i)=-1/sqrt2+1/sqrt2i$

Explanation:

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

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

$e^{\frac{π}{4} i} \cdot e^{\frac{π}{2} i}$ $e^(pi/4i)*e^(pi/2i)$

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

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

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

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

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

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

= $- \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i$ $-1/sqrt2+1/sqrt2i$

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