How can you use trigonometric functions to simplify $4 e^{\frac{5 π}{4} i}$ $4 e^( ( 5 pi)/4 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $4 e^{\frac{5 π}{4} i}$ $4 e^( ( 5 pi)/4 i )$ into a non-exponential complex number?

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Answer 1 · 2015-12-21T12:03:43

Use the Moivre formula.

Explanation:

The Moivre formula tells us that $e^{i θ} = cos (θ) + i sin (θ)$ $e^(itheta) = cos(theta) + isin(theta)$ .

Apply this here : $4 e^{i \frac{5 π}{4}} = 4 (cos (\frac{5 π}{4}) + i sin (\frac{5 π}{4}))$ $4e^(i(5pi)/4) = 4(cos((5pi)/4) + isin((5pi)/4))$

On the trigonometric circle, $\frac{5 π}{4} = \frac{- 3 π}{4}$ $(5pi)/4 = (-3pi)/4$ . Knowing that $cos (\frac{- 3 π}{4}) = - \frac{\sqrt{2}}{2}$ $cos((-3pi)/4) = -sqrt2/2$ and $sin (\frac{- 3 π}{4}) = - \frac{\sqrt{2}}{2}$ $sin((-3pi)/4) = -sqrt2/2$ , we can say that $4 e^{i \frac{5 π}{4}} = 4 (- \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}) = - 2 \sqrt{2} - 2 i \sqrt{2}$ $4e^(i(5pi)/4) = 4(-sqrt2/2 -i(sqrt2)/2) = -2sqrt2 -2isqrt2$ .

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How can you use trigonometric functions to simplify $4 e^{\frac{5 π}{4} i}$ $4 e^( ( 5 pi)/4 i )$ into a non-exponential complex number?

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Explanation:

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How can you use trigonometric functions to simplify 4 e^( ( 5 pi)/4 i ) 4e5π4i 4 e^( ( 5 pi)/4 i ) into a non-exponential complex number?

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How can you use trigonometric functions to simplify $4 e^{\frac{5 π}{4} i}$ $4 e^( ( 5 pi)/4 i )$ into a non-exponential complex number?