How do you multiply $e^{\frac{7 π}{4} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 7 pi )/ 4 i) * e^( 3 pi/2 i )$ in trigonometric form?

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How do you multiply $e^{\frac{7 π}{4} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 7 pi )/ 4 i) * e^( 3 pi/2 i )$ in trigonometric form?

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Answer 1 · 2016-03-02T10:14:33

$e^{7 \frac{π}{4} i} = (cos (7 \frac{π}{4}) + i sin (7 \frac{π}{4})) = \frac{1}{\sqrt{2}} - i (\frac{1}{\sqrt{2}})$ $e^(7pi/4i)=(cos(7pi/4)+isin(7pi/4))=1/sqrt2-i(1/sqrt2)$

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

Explanation:

Trigonometric form of $e^{7 \frac{π}{4} i}$ $e^(7pi/4i)$ can be written as

$(cos (7 \frac{π}{4}) + i sin (7 \frac{π}{4}))$ $(cos(7pi/4)+isin(7pi/4))$

As $cos (7 \frac{π}{4}) = cos (- \frac{π}{4}) = cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ $cos(7pi/4)=cos(-pi/4)=cos(pi/4)=1/sqrt2$ and $sin (7 \frac{π}{4}) = sin (- \frac{π}{4}) = sin (\frac{π}{4}) = - \frac{1}{\sqrt{2}}$ $sin(7pi/4)=sin(-pi/4)=sin(pi/4)=-1/sqrt2$

$e^{7 \frac{π}{4} i}$ $e^(7pi/4i)$ can be written as $\frac{1}{\sqrt{2}} - i (\frac{1}{\sqrt{2}})$ $1/sqrt2-i(1/sqrt2)$

and that of $e^{3 \frac{π}{2} i}$ $e^(3pi/2i)$ can be written as

$(cos (3 \frac{π}{2}) + i sin (3 \frac{π}{2}))$ $(cos(3pi/2)+isin(3pi/2))$ and as $cos (3 \frac{π}{2}) = 0$ $cos(3pi/2)=0$ and $sin (\frac{π}{2}) = - 1$ $sin(pi/2)=-1$ ,

$e^{3 \frac{π}{2} i}$ $e^(3pi/2i)$ can be written as $- i$ $-i$

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How do you multiply $e^{\frac{7 π}{4} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 7 pi )/ 4 i) * e^( 3 pi/2 i )$ in trigonometric form?

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How do you multiply $e^{\frac{7 π}{4} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 7 pi )/ 4 i) * e^( 3 pi/2 i )$ in trigonometric form?