How can you use trigonometric functions to simplify $9 e^{\frac{7 π}{6} i}$ $9 e^( ( 7 pi)/6 i )$ into a non-exponential complex number?

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Answer 1 · 2017-02-01T06:46:35

The answer is $= - \frac{9}{2} (\sqrt{3} + i)$ $=-9/2(sqrt3+i)$

We need Euler's formula

$e^{i θ} = cos θ + i sin θ$ $e^(itheta)=costheta+isintheta$

Therefore,

$9 e^{\frac{7}{6} π i} = 9 (cos (\frac{7}{6} π) + i sin (\frac{7}{6} π))$ $9e^(7/6pi i)=9(cos(7/6pi)+isin(7/6pi))$

$cos (\frac{7}{6} π) = - \frac{\sqrt{3}}{2}$ $cos(7/6pi)=-sqrt3/2$

$sin (\frac{7}{6} π) = - \frac{1}{2}$ $sin(7/6pi)=-1/2$

So,

$9 e^{\frac{7}{6} π i} = 9 (- \frac{\sqrt{3}}{2} - \frac{i}{2})$ $9e^(7/6pi i)=9(-sqrt3/2-i/2)$

$= - \frac{9}{2} (\sqrt{3} + i)$ $=-9/2(sqrt3+i)$

How can you use trigonometric functions to simplify 9e7π6i 9 e^( ( 7 pi)/6 i ) into a non-exponential complex number?