How can you use trigonometric functions to simplify $14 e^{\frac{π}{8} i}$ $14 e^( ( pi)/8 i )$ into a non-exponential complex number?

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Answer 1 · 2016-10-27T05:46:12

The complex number is $14 (cos (\frac{π}{8}) + i sin (\frac{π}{8}))$ $14(cos(pi/8)+isin(pi/8))$

Let $z$ $z$ be the complex number

So $z = 14 e^{i \frac{π}{8}}$ $z=14e^(ipi/8)$

We use $z = r e^{i θ} = r (cos θ + i sin θ)$ $z=re^(itheta)=r(costheta+isintheta)$

So $r = 14$ $r=14$ and $θ = \frac{π}{8}$ $theta=pi/8$

$z = 14 (cos (\frac{π}{8}) + i sin (\frac{π}{8}))$ $z=14(cos(pi/8)+isin(pi/8))$

How can you use trigonometric functions to simplify 14eπ8i 14 e^( ( pi)/8 i ) into a non-exponential complex number?