How do you multiply $e^{\frac{5 π}{4} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 5 pi )/ 4 i) * e^( 3 pi/2 i )$ in trigonometric form?

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How do you multiply $e^{\frac{5 π}{4} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( 5 pi )/ 4 i) * e^( 3 pi/2 i )$ in trigonometric form?

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Answer 1 · 2018-04-11T09:00:52

Using the formula $e^{i θ} = cos (θ) + i sin (θ)$ $e^(itheta) = cos(theta) + isin(theta)$ and some trigonometric identities.

Explanation:

The formula:

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

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

Multiplication:

$e^{\frac{5 π}{4} i} \times e^{\frac{3 π}{2} i} = (cos (\frac{5 π}{4}) + i sin (\frac{5 π}{4})) (cos (\frac{3 π}{2}) + i sin (\frac{3 π}{2}))$ $e^((5pi)/4i) xxe^((3pi)/2i) = (cos((5pi)/4) + isin((5pi)/4))(cos((3pi)/2) + isin((3pi)/2))$

$= cos (\frac{5 π}{4}) cos (\frac{3 π}{2}) + i cos (\frac{5 π}{4}) sin (\frac{3 π}{2}) + i sin (\frac{5 π}{4}) cos (\frac{3 π}{2}) - sin (\frac{5 π}{4}) sin (\frac{3 π}{2})$ $= cos((5pi)/4)cos((3pi)/2)+icos((5pi)/4)sin((3pi)/2) + isin((5pi)/4)cos((3pi)/2) -sin((5pi)/4)sin((3pi)/2)$

$= cos (\frac{5 π}{4}) cos (\frac{3 π}{2}) - sin (\frac{5 π}{4}) sin (\frac{3 π}{2}) + i (cos (\frac{5 π}{4}) sin (\frac{3 π}{2}) + sin (\frac{5 π}{4}) cos (\frac{3 π}{2}))$ $= cos((5pi)/4)cos((3pi)/2)-sin((5pi)/4)sin((3pi)/2) + i(cos((5pi)/4)sin((3pi)/2) + sin((5pi)/4)cos((3pi)/2))$

$= cos (\frac{5 π}{4} + \frac{3 π}{2}) + i sin (\frac{5 π}{4} + \frac{3 π}{2})$ $= cos((5pi)/4 + (3pi)/2) + isin((5pi)/4 + (3pi)/2)$

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

$= e^{\frac{11 π}{4} i}$ $= e^((11pi)/4i)$

Its much easier if you do this:

$e^{\frac{5 π}{4} i} \times e^{\frac{3 π}{2} i} = e^{\frac{5 π}{4} i + \frac{3 π}{2} i}$ $e^((5pi)/4i) xxe^((3pi)/2i) = e^((5pi)/4i + (3pi)/2i)$

$= e^{\frac{11 π}{4} i}$ $= e^((11pi)/4i)$

Answer 2 · 2018-04-11T09:08:58

The answer is $= - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i$ $=-sqrt2/2+sqrt2/2i$

Explanation:

Apply Eulers' Identity

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

Therefore,

$e^{\frac{5}{4} π i} \cdot e^{\frac{3}{2} π i} = (cos (\frac{5}{4} π) + i sin (\frac{5}{4} π)) (cos (\frac{3}{2} π) + i sin (\frac{3}{2} π))$ $e^(5/4pii)*e^(3/2pii)=(cos(5/4pi)+isin(5/4pi))(cos(3/2pi)+isin(3/2pi))$

$cos (\frac{5}{4} π) = cos (\frac{1}{4} π + π) = cos (\frac{1}{4} π) cos (π) - sin (\frac{1}{4} π) sin (π)$ $cos(5/4pi)=cos(1/4pi+pi)=cos(1/4pi)cos(pi)-sin(1/4pi)sin(pi)$

$= - \frac{\sqrt{2}}{2}$ $=-sqrt2/2$

$sin (\frac{5}{4} π) = sin (\frac{1}{4} π + π) = sin (\frac{1}{4} π) cos (π) + cos (\frac{1}{4} π) sin (π)$ $sin(5/4pi)=sin(1/4pi+pi)=sin(1/4pi)cos(pi)+cos(1/4pi)sin(pi)$

$= - \frac{\sqrt{2}}{2}$ $=-sqrt2/2$

$cos (\frac{3}{2} π) = cos (\frac{1}{2} π + π) = cos (\frac{1}{2} π) cos (π) - sin (\frac{1}{2} π) sin (π)$ $cos(3/2pi)=cos(1/2pi+pi)=cos(1/2pi)cos(pi)-sin(1/2pi)sin(pi)$

$= 0$ $=0$

$sin (\frac{3}{2} π) = sin (\frac{1}{2} π + π) = sin (\frac{1}{2} π) cos (π) + cos (\frac{1}{2} π) sin (π)$ $sin(3/2pi)=sin(1/2pi+pi)=sin(1/2pi)cos(pi)+cos(1/2pi)sin(pi)$

$= - 1$ $=-1$

Finally,

$e^{\frac{5}{4} π i} \cdot e^{\frac{3}{2} π i} = (- \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i) (- i)$ $e^(5/4pii)*e^(3/2pii)=(-sqrt2/2-sqrt2/2i)(-i)$

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

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