How can you use trigonometric functions to simplify $11 e^{\frac{4 π}{3} i}$ $11 e^( ( 4 pi)/3 i )$ into a non-exponential complex number?

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Answer 1 · 2016-03-28T09:07:27

$e^{\frac{4 π}{3} i} = - \frac{\sqrt{3}}{2} - \frac{1}{2} i$ $e^((4pi)/3i)=-sqrt3/2-1/2i$

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

$e^{\frac{4 π}{3} i} = cos (\frac{4 π}{3}) + i sin (\frac{4 π}{3})$ $e^((4pi)/3i)=cos((4pi)/3)+isin((4pi)/3)$

= $cos (π + \frac{π}{3}) + i sin (π + \frac{π}{3})$ $cos(pi+(pi)/3)+isin(pi+(pi)/3)$

= $- cos (\frac{π}{3}) + i (- sin (\frac{π}{3}))$ $-cos((pi)/3)+i(-sin((pi)/3))$

= $- \frac{\sqrt{3}}{2} - \frac{1}{2} i$ $-sqrt3/2-1/2i$

How can you use trigonometric functions to simplify 11 e^( ( 4 pi)/3 i ) 11e4π3i 11 e^( ( 4 pi)/3 i ) into a non-exponential complex number?