How can you use trigonometric functions to simplify $5 e^{\frac{3 π}{2} i}$ $5 e^( ( 3 pi)/2 i )$ into a non-exponential complex number?

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Answer 1 · 2015-12-29T00:07:16

Apply Euler's formula to convert the function into a trigonometric form, and then evaluate to find

$5 e^{\frac{3 π}{2} i} = - 5 i$ $5e^((3pi)/2i) = -5i$

Using Euler's formula: $e^{i θ} = cos (θ) + i sin (θ)$ $e^(itheta) = cos(theta)+isin(theta)$

$5 e^{i \frac{3 π}{2}} = 5 (cos (\frac{3 π}{2}) + i sin (\frac{3 π}{2}))$ $5e^(i(3pi)/2) = 5(cos((3pi)/2) + isin((3pi)/2))$

$= 5 (0 + i (- 1))$ $= 5(0 + i(-1))$

$= - 5 i$ $= -5i$

How can you use trigonometric functions to simplify 5 e^( ( 3 pi)/2 i ) 5e3π2i 5 e^( ( 3 pi)/2 i ) into a non-exponential complex number?