How can you use trigonometric functions to simplify $6 e^{\frac{π}{8} i}$ $6 e^( ( pi)/8 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $6 e^{\frac{π}{8} i}$ $6 e^( ( pi)/8 i )$ into a non-exponential complex number?

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Answer 1 · 2016-03-25T12:41:12

$6 e^{i \frac{π}{8}} = 6 (cos (\frac{π}{8}) + i sin (\frac{π}{8}))$ $6 e^{i pi/8} = 6 (cos(pi/8) + i sin(pi/8))$

$= 3 \sqrt{2 + \sqrt{2}} + i (3 \sqrt{2 - \sqrt{2}})$ $= 3sqrt{2+sqrt2} + i(3sqrt{2-sqrt2})$

Explanation:

Euler showed that for a complex number $z = r e^{i θ}$ $z = r e^{i theta}$ ,

$r e^{i θ} = r (cos (θ) + i sin (θ))$ $r e^{i theta} = r (cos(theta) + i sin(theta))$

In this case

$r = 6$ $r = 6$
$θ = \frac{π}{8}$ $theta = pi/8$

Therefore,

$6 e^{i \frac{π}{8}} = 6 (cos (\frac{π}{8}) + i sin (\frac{π}{8}))$ $6 e^{i pi/8} = 6 (cos(pi/8) + i sin(pi/8))$

You can use half angle formulas to find $sin (\frac{π}{8})$ $sin(pi/8)$ and $cos (\frac{π}{8})$ $cos(pi/8)$ .

$6 (cos (\frac{π}{8}) + i sin (\frac{π}{8})) = 6 (\frac{\sqrt{2 + \sqrt{2}}}{2} + i (\frac{\sqrt{2 - \sqrt{2}}}{2}))$ $6 (cos(pi/8) + i sin(pi/8)) = 6(sqrt{2+sqrt2}/2 + i(sqrt{2-sqrt2}/2))$

$= 3 \sqrt{2 + \sqrt{2}} + i (3 \sqrt{2 - \sqrt{2}})$ $= 3sqrt{2+sqrt2} + i(3sqrt{2-sqrt2})$

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How can you use trigonometric functions to simplify $6 e^{\frac{π}{8} i}$ $6 e^( ( pi)/8 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $6 e^{\frac{π}{8} i}$ $6 e^( ( pi)/8 i )$ into a non-exponential complex number?