How can you use trigonometric functions to simplify $16 e^{\frac{π}{6} i}$ $16 e^( ( pi)/6 i )$ into a non-exponential complex number?

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Answer 1 · 2015-12-31T10:23:42

Euler formula $e^{i θ} = cos (θ) + i sin (θ)$ $e^(i theta) = cos(theta)+isin(theta)$ using this we can represent the given complex number as .

$16 e^{\frac{π}{6} i} = 8 \sqrt{3} + 8 i$ $16e^(pi/6i) = 8sqrt(3)+8i$

$16 e^{\frac{π}{6} i}$ $16e^(pi/6i)$
Comparing to the Euler's formula $e^{i θ}$ $e^(i theta)$
We can see $θ = \frac{π}{6}$ $theta=pi/6$

Therefore,

$16 e^{\frac{π}{6} i} = 16 (cos (\frac{π}{6}) + i sin (\frac{π}{6}))$ $16e^(pi/6 i) = 16(cos(pi/6) + isin(pi/6))$

$16 e^{\frac{π}{6} i} = 16 (\sqrt{\frac{3}{2}} + i (\frac{1}{2}))$ $16e^(pi/6i) = 16(sqrt(3/2) + i (1/2))$

$16 e^{\frac{π}{6} i} = \frac{16}{2} (\sqrt{3} + i)$ $16e^(pi/6i) = 16/2(sqrt(3) + i)$

$16 e^{\frac{π}{6} i} = 8 (\sqrt{3} + i)$ $16e^(pi/6i) = 8(sqrt(3)+i)$

$16 e^{\frac{π}{6} i} = 8 \sqrt{3} + 8 i$ $16e^(pi/6i) = 8sqrt(3)+8i$ Answer

How can you use trigonometric functions to simplify 16eπ6i 16 e^( ( pi)/6 i ) into a non-exponential complex number?