How can you use trigonometric functions to simplify $8 e^{\frac{7 π}{4} i}$ $8 e^( ( 7 pi)/4 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $8 e^{\frac{7 π}{4} i}$ $8 e^( ( 7 pi)/4 i )$ into a non-exponential complex number?

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Answer 1 · 2015-12-28T18:51:02

$8 \cdot e^{\frac{7 π}{4} \cdot i} = 4 π - 4 π \cdot i$ $8*e^((7pi)/4*i)=4pi-4pi*i$

Explanation:

If we have a complex number in trigonometric form:

$z = | z | \cdot e^{i \cdot φ}$ $z=|z|*e^(i*varphi)$ ,

then according to de Moivre formula we can write, that:

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

If we assume that the module $| z |$ $|z|$ is equal to $1$ $1$ , then the formula simplifies to:

$e^{i \cdot φ} = (cos φ + i \cdot sin φ)$ $e^(i*varphi)=(cosvarphi+i*sinvarphi)$

So the expression $8 \cdot e^{\frac{7 π}{4} \cdot i}$ $8*e^((7pi)/4*i)$ can be written as:

$8 \cdot (cos (\frac{7 π}{4}) + i \cdot sin (\frac{7 π}{4})) =$ $8*(cos((7pi)/4)+i*sin((7pi)/4))=$

$8 \cdot [cos (2 π - \frac{π}{4}) + i \cdot sin (2 π - \frac{π}{4})] =$ $8*[cos(2pi-pi/4)+i*sin(2pi-pi/4)]=$

$8 \cdot [cos (\frac{π}{4}) - i \cdot sin (\frac{π}{4})] = 8 \cdot [\frac{\sqrt{2}}{2} - i \cdot \frac{\sqrt{2}}{2}] =$ $8*[cos(pi/4)-i*sin(pi/4)]=8*[sqrt(2)/2-i*sqrt(2)/2]=$

$4 \sqrt{2} - 4 \sqrt{2} \cdot i$ $4sqrt(2)-4sqrt(2)*i$

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How can you use trigonometric functions to simplify $8 e^{\frac{7 π}{4} i}$ $8 e^( ( 7 pi)/4 i )$ into a non-exponential complex number?

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Explanation:

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How can you use trigonometric functions to simplify 8e7π4i 8 e^( ( 7 pi)/4 i ) into a non-exponential complex number?

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How can you use trigonometric functions to simplify $8 e^{\frac{7 π}{4} i}$ $8 e^( ( 7 pi)/4 i )$ into a non-exponential complex number?