How can you use trigonometric functions to simplify $13 e^{\frac{π}{8} i}$ $13 e^( ( pi)/8 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $13 e^{\frac{π}{8} i}$ $13 e^( ( pi)/8 i )$ into a non-exponential complex number?

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Answer 1 · 2016-03-08T19:20:16

$13 [cos (\frac{π}{8}) + i sin (\frac{π}{8})]$ $13 [ cos(pi/8) + isin(pi/8) ]$

Explanation:

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

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

here r = 13 and $θ = \frac{π}{8}$ $theta = pi/8$

$\Rightarrow 13 e^{\frac{π}{8} i} = 13 [cos (\frac{π}{8}) + i sin (\frac{π}{8})]$ $rArr 13e^(pi/8 i) = 13 [ cos(pi/8) + isin(pi/8) ]$

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How can you use trigonometric functions to simplify $13 e^{\frac{π}{8} i}$ $13 e^( ( pi)/8 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $13 e^{\frac{π}{8} i}$ $13 e^( ( pi)/8 i )$ into a non-exponential complex number?