How can you use trigonometric functions to simplify $15 e^{\frac{2 π}{3} i}$ $15 e^( ( 2 pi)/3 i )$ into a non-exponential complex number?

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Answer 1 · 2017-11-19T13:27:05

The answer is $= \frac{15}{2} (- 1 + i \sqrt{3})$ $=15/2(-1+isqrt3)$

Apply Euler's formula

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

Therefore,

$15 e^{\frac{2}{3} π i} = 15 (cos (\frac{2}{3} π) + i sin (\frac{2}{3} π))$ $15e^(2/3pii)=15(cos(2/3pi)+isin(2/3pi))$

$cos (\frac{2}{3} π) = - \frac{1}{2}$ $cos(2/3pi)=-1/2$

$sin (\frac{2}{3} π) = \frac{\sqrt{3}}{2}$ $sin(2/3pi)=sqrt3/2$

So,

$15 e^{\frac{2}{3} π i} = 15 (- \frac{1}{2} + \frac{\sqrt{3}}{2} i)$ $15e^(2/3pii)=15(-1/2+sqrt3/2i)$

$= \frac{15}{2} (- 1 + i \sqrt{3})$ $=15/2(-1+isqrt3)$

How can you use trigonometric functions to simplify 15e2π3i 15 e^( ( 2 pi)/3 i ) into a non-exponential complex number?