How can you use trigonometric functions to simplify $9 e^{\frac{5 π}{3} i}$ $9 e^( ( 5 pi)/3 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $9 e^{\frac{5 π}{3} i}$ $9 e^( ( 5 pi)/3 i )$ into a non-exponential complex number?

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Answer 1 · 2016-03-05T21:14:10

$9 [cos (\frac{5 π}{3}) + i sin (\frac{5 π}{3})]$ $9[cos((5pi)/3) + isin((5pi)/3) ]$

Explanation:

Using $Euler's relationship$ $color(blue)" Euler's relationship "$

which states $r e^{θ i} = r (cos θ + i sin θ)$ $re^(theta i) = r(costheta + isintheta)$

here r = 9 and $θ = \frac{5 π}{3}$ $theta = (5pi)/3$

$\Rightarrow 9 e^{\frac{5 π}{3} i} = 9 [cos (\frac{5 π}{3}) + i sin (\frac{5 π}{3})]$ $rArr9e^((5pi)/3 i) = 9 [cos((5pi)/3) + isin((5pi)/3) ]$

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How can you use trigonometric functions to simplify $9 e^{\frac{5 π}{3} i}$ $9 e^( ( 5 pi)/3 i )$ into a non-exponential complex number?

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Science

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How can you use trigonometric functions to simplify 9e5π3i 9 e^( ( 5 pi)/3 i ) into a non-exponential complex number?

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Explanation:

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How can you use trigonometric functions to simplify $9 e^{\frac{5 π}{3} i}$ $9 e^( ( 5 pi)/3 i )$ into a non-exponential complex number?