How can you use trigonometric functions to simplify $27 e^{\frac{π}{2} i}$ $27 e^( ( pi)/2 i )$ into a non-exponential complex number?

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Answer 1 · 2015-12-24T09:14:02

Use the equation $e^{i θ} = cos (θ) + i sin (θ)$ $e^(itheta) = cos(theta)+isin(theta)$ to find
$27 e^{\frac{π}{2} i} = 27 i$ $27e^(pi/2i) = 27i$

Using the identity $e^{i θ} = cos (θ) + i sin (θ)$ $e^(itheta) = cos(theta) + isin(theta)$ we have

$27 e^{\frac{π}{2} i} = 27 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))$ $27e^(pi/2i) = 27(cos(pi/2)+isin(pi/2))$

$= 27 (0 + i (1))$ $= 27(0 + i(1))$

$= 27 i$ $= 27i$

How can you use trigonometric functions to simplify 27eπ2i 27 e^( ( pi)/2 i ) into a non-exponential complex number?