How can you use trigonometric functions to simplify $23 e^{\frac{π}{8} i}$ $23 e^( ( pi)/8 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $23 e^{\frac{π}{8} i}$ $23 e^( ( pi)/8 i )$ into a non-exponential complex number?

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Answer 1 · 2016-04-08T15:11:50

$23 e^{\frac{π}{8} i} = - 20.08 - 11.2 i$ $23e^(pi/8i) = - 20.08 - 11.2i$

Explanation:

According to Euler's formula,

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

When $\frac{π}{8}$ $pi/8$ is substituted as $x$ $x$ , then

$e^{\frac{π}{8} i} = cos (\frac{π}{8}) + i sin (\frac{π}{8})$ $e^(pi/8i) = cos(pi/8) + isin(pi/8)$
$= cos ({22.5}^{o}) + i sin ({22.5}^{o})$ $= cos(22.5^o) + isin(22.5^o)$
$= - 0.873 - 0.487 i$ $= -0.873 -0.487i$ ,

which is the value of $e^{\frac{π}{8} i}$ $e^(pi/8i)$ .

Multiply the whole thing by $23$ $23$ to find the entire thing:

$23 (- 0.873 - 0.487 i) = - 20.08 - 11.2 i$ $23(- 0.873 - 0.487i) = - 20.08 - 11.2i$

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How can you use trigonometric functions to simplify $23 e^{\frac{π}{8} i}$ $23 e^( ( pi)/8 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify 23eπ8i 23 e^( ( pi)/8 i ) into a non-exponential complex number?

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Explanation:

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How can you use trigonometric functions to simplify $23 e^{\frac{π}{8} i}$ $23 e^( ( pi)/8 i )$ into a non-exponential complex number?