How do you multiply $e^{\frac{5 π}{4} i} \cdot e^{\frac{π}{2} i}$ $e^(( 5 pi )/ 4 i) * e^( pi/2 i )$ in trigonometric form?

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How do you multiply $e^{\frac{5 π}{4} i} \cdot e^{\frac{π}{2} i}$ $e^(( 5 pi )/ 4 i) * e^( pi/2 i )$ in trigonometric form?

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Answer 1 · 2016-04-13T12:35:50

$e^{(5 \frac{π}{4}) i} \cdot e^{(\frac{π}{2}) i} = - \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}$ $e^((5pi/4)i)*e^((pi/2)i)=-1/sqrt2-i1/sqrt2$

Explanation:

A complex number can be written in polar form in two ways - either as $r \cdot e^{i θ}$ $r*e^(itheta)$ or as $r cos θ + i r sin θ$ $rcostheta+irsintheta$ .

Hence,

$e^{(5 \frac{π}{4}) i} = cos (5 \frac{π}{4}) + i sin (5 \frac{π}{4})$ $e^((5pi/4)i)=cos(5pi/4)+isin(5pi/4)$ and

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

Hence, $e^{(5 \frac{π}{4}) i} \cdot e^{(\frac{π}{2}) i}$ $e^((5pi/4)i)*e^((pi/2)i)$

= $(cos (5 \frac{π}{4}) + i sin (5 \frac{π}{4})) \cdot (cos (\frac{π}{2}) + i sin (\frac{π}{2}))$ $(cos(5pi/4)+isin(5pi/4))*(cos(pi/2)+isin(pi/2))$

= $(cos (5 \frac{π}{4}) + i sin (5 \frac{π}{4})) \cdot (0 + i)$ $(cos(5pi/4)+isin(5pi/4))*(0+i)$

as $cos (\frac{π}{2}) = 0$ $cos(pi/2)=0$ and $sin (\frac{π}{2}) = 1$ $sin(pi/2)=1$

= $(i cos (5 \frac{π}{4}) + i^{2} sin (5 \frac{π}{4}))$ $(icos(5pi/4)+i^2sin(5pi/4))$

= $- sin (5 \frac{π}{4}) + i cos (5 \frac{π}{4})$ $-sin(5pi/4)+icos(5pi/4)$

= $- \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}$ $-1/sqrt2-i1/sqrt2$

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How do you multiply $e^{\frac{5 π}{4} i} \cdot e^{\frac{π}{2} i}$ $e^(( 5 pi )/ 4 i) * e^( pi/2 i )$ in trigonometric form?

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Explanation:

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