How can you use trigonometric functions to simplify $8 e^{\frac{3 π}{2} i}$ $8 e^( ( 3 pi)/2 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $8 e^{\frac{3 π}{2} i}$ $8 e^( ( 3 pi)/2 i )$ into a non-exponential complex number?

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Answer 1 · 2016-04-16T11:13:07

$- 8 i$ $-8 i$

Explanation:

$e^{i x} = cos x + i sin x$ $e^(ix) = cos x + i sin x$

So, $8 e^{\frac{3 π}{2} i} = 8 (cos (\frac{3 π}{2}) + i sin (\frac{3 π}{2}))$ $8 e^((3pi)/2i)=8(cos((3pi)/2) + i sin ((3pi)/2))$
$cos (\frac{3 π}{2}) = 0 and sin (\frac{3 π}{2}) = sin (π + \frac{π}{2}) = - sin (\frac{π}{2}) = - 1$ $cos((3pi)/2) = 0 and sin ((3pi)/2)=sin(pi+pi/2)=-sin(pi/2)=-1$

So, $8 e^{\frac{3 π}{2} i} = - 8 i$ $8 e^((3pi)/2i)= -8 i$ .

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How can you use trigonometric functions to simplify $8 e^{\frac{3 π}{2} i}$ $8 e^( ( 3 pi)/2 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $8 e^{\frac{3 π}{2} i}$ $8 e^( ( 3 pi)/2 i )$ into a non-exponential complex number?