How can you use trigonometric functions to simplify $9 e^{\frac{π}{3} i}$ $9 e^( ( pi)/3 i )$ into a non-exponential complex number?

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Answer 1 · 2016-07-31T18:33:34

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

We recall that, $e^{i θ} = cos θ + i sin θ$ $e^(itheta)=costheta+isintheta$ .

Hence, $9 e^{i \frac{π}{3}} = 9 (cos (\frac{π}{3}) + i sin (\frac{π}{3}))$ $9e^(ipi/3)=9(cos(pi/3)+isin(pi/3))$

$= 9 (\frac{1}{2} + i \frac{\sqrt{3}}{2})$ $=9(1/2+isqrt3/2)$

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

How can you use trigonometric functions to simplify 9eπ3i 9 e^( ( pi)/3 i ) into a non-exponential complex number?