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 · 2016-06-14T20:37:29

The value of this complex number is $- 2 \sqrt{2} - 2 \sqrt{2} i$ $-2sqrt(2)-2sqrt(2)i$

Explanation:

To express a complex number without exponents you use the following formula:

$| z | \cdot e^{φ i} = | z | (cos φ + i sin φ)$ $|z|*e^(varphii)=|z|(cosvarphi+isinvarphi)$

In the example above you have:

$| z | = 4$ $|z|=4$

$φ = \frac{5 π}{4}$ $varphi=(5pi)/4$

So you have:

$4 e^{\frac{5 π}{4} \cdot i} = 4 \cdot (cos (\frac{5 π}{4}) + i \cdot sin (\frac{5 π}{4})) =$ $4e^((5pi)/4*i)=4*(cos((5pi)/4)+i*sin((5pi)/4))=$

$4 \cdot (cos (π + \frac{π}{4}) + i \cdot sin (π + \frac{π}{4})) =$ $4*(cos(pi+pi/4)+i*sin(pi+pi/4))=$
$4 \cdot (- cos (\frac{π}{4}) - i sin (\frac{π}{4})) =$ $4*(-cos(pi/4)-isin(pi/4))=$
$4 \cdot (- \frac{\sqrt{2}}{2} - i \cdot \frac{\sqrt{2}}{2}) = - 2 \sqrt{2} - 2 \sqrt{2} \cdot i$ $4*(-sqrt(2)/2-i*sqrt(2)/2)=-2sqrt(2)-2sqrt(2)*i$

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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?