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

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Answer 1 · 2018-03-25T08:05:58

The answer is $= 1 + i \sqrt{3}$ $=1+isqrt3$

Apply Euler's Identity

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

Here,

$z = 2 e^{i \frac{π}{3}}$ $z=2e^(ipi/3)$

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

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

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

Therefore,

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

$= 1 + i \sqrt{3}$ $=1+isqrt3$

How can you use trigonometric functions to simplify 2 e^( ( pi)/3 i ) 2eπ3i 2 e^( ( pi)/3 i ) into a non-exponential complex number?