How can you use trigonometric functions to simplify $18 e^{\frac{5 π}{3} i}$ $18 e^( ( 5 pi)/3 i )$ into a non-exponential complex number?

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Answer 1 · 2016-12-13T13:00:26

The answer is $= 9 (1 - i \sqrt{3})$ $=9(1-isqrt3)$

We use

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

$cos (\frac{5}{3} π) = \frac{1}{2}$ $cos(5/3pi)=1/2$

$sin (\frac{5}{3} π) = - \frac{\sqrt{3}}{2}$ $sin(5/3pi)=-sqrt3/2$

$18 e^{\frac{5}{3} i π} = 18 (cos (\frac{5}{3} π) + i sin (\frac{5}{3} π))$ $18e^(5/3ipi)=18(cos(5/3pi)+isin(5/3pi))$

$= 18 (\frac{1}{2} - i \frac{\sqrt{3}}{2})$ $=18(1/2-isqrt3/2)$

$= 9 (1 - i \sqrt{3})$ $=9(1-isqrt3)$

How can you use trigonometric functions to simplify 18 e^( ( 5 pi)/3 i ) 18e5π3i 18 e^( ( 5 pi)/3 i ) into a non-exponential complex number?