How do you multiply $e^{\frac{3 π}{8} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 3 pi )/ 8 i) * e^( 3 pi/2 i )$ in trigonometric form?

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How do you multiply $e^{\frac{3 π}{8} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 3 pi )/ 8 i) * e^( 3 pi/2 i )$ in trigonometric form?

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Answer 1 · 2017-10-15T19:42:32

$= e^{(\frac{31 π}{8}) i}$ $= e^(((31pi)/8)i)$

Explanation:

$e^{(\frac{3 π}{8}) i} \cdot e^{3 (\frac{π}{2}) i}$ $e^(((3pi)/8)i) * e^ (3(pi/2)i$
$e^{(\frac{3 π}{8}) i} \cdot e^{(\frac{7 π}{2}) i}$ $e^(((3pi)/8)i) * e^(((7pi)/2)i)$

$e^{(\frac{3 π}{8}) i} \cdot e^{(\frac{28 π}{8}) i}$ $e^(((3pi)/8)i)* e^ (((28pi)/8)i)$
$e^{(\frac{3 π + 28 π}{8}) (i + i) = e^{(\frac{31 π}{8}) i}}$ $e^ (((3pi + 28pi)/8)(i+i) = e^(((31pi)/8)i)$

Answer 2 · 2017-10-15T20:53:31

$cos (\frac{π}{8}) - i sin (\frac{π}{8})$ $cos(pi/8)-isin(pi/8)$ or $- {(- 1)}^{\frac{7}{8}}$ $-(-1)^(7/8)$

Explanation:

Are you sure its $e^{3 \frac{π}{2}}$ $e^(3pi/2)$ not $e^{\frac{3 π}{2}}$ $e^((3pi)/2)$ because it makes more sense when you write it in trig form.

Remember this beauty?
$e^{i π} = - 1$ $e^(ipi)=-1$
That was Euler's identity. This is the generalized formula:
$e^{i x} = cos x + i sin x$ $e^(ix)=cosx+isinx$

Therefore, we can break down the two terms involved:

$e^{i \frac{3}{8} π} = cos (\frac{3 π}{8}) + i sin (\frac{3 π}{8})$ $e^(i3/8pi)=cos((3pi)/8)+isin((3pi)/8)$

$e^{i \frac{7}{2} π} = cos (\frac{7 π}{2}) + i sin (\frac{7 π}{2}) = 0 + i \cdot (- 1) = - i$ $e^(i7/2pi)=cos((7pi)/2)+isin((7pi)/2)=0+i*(-1)=-i$

$e^{i \frac{3}{8} π} = {(e^{i \frac{3}{2} π})}^{\frac{1}{4}} = {(- i)}^{\frac{1}{4}}$ $e^(i3/8pi)=(e^(i3/2pi))^(1/4)=(-i)^(1/4)$
$e^{i \frac{3}{8} π} \cdot e^{i \frac{3}{2} π} = {(e^{i \frac{3}{2} π})}^{\frac{1}{4}} \cdot e^{i \frac{7}{2} π} = {(e^{i \frac{3}{2} π})}^{\frac{5}{4}} = {(- i)}^{\frac{5}{4}} = - {(- 1)}^{\frac{7}{8}}$ $e^(i3/8pi)*e^(i3/2pi)=(e^(i3/2pi))^(1/4)*e^(i7/2pi)=(e^(i3/2pi))^(5/4)=(-i)^(5/4)=-(-1)^(7/8)$

Or

$e^{i \frac{3}{8} π} \cdot e^{i \frac{7}{2} π} = sin (\frac{3 π}{8}) - i cos (\frac{3 π}{8})$ $e^(i3/8pi)*e^(i7/2pi) = sin((3pi)/8)-icos((3pi)/8)$

$= cos (\frac{π}{8}) - i sin (\frac{π}{8})$ $=cos(pi/8)-isin(pi/8)$

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How do you multiply $e^{\frac{3 π}{8} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 3 pi )/ 8 i) * e^( 3 pi/2 i )$ in trigonometric form?

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How do you multiply $e^{\frac{3 π}{8} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 3 pi )/ 8 i) * e^( 3 pi/2 i )$ in trigonometric form?