How can you use trigonometric functions to simplify $2 e^{\frac{π}{8} i}$ $2 e^( ( pi)/8 i )$ into a non-exponential complex number?

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Answer 1 · 2016-04-11T05:02:03

$1.8478 + 0.7654 i$ $1.8478+0.7654i$

We can use the identity $r e^{i θ} = r cos θ + i r sin θ$ $re^(itheta)=rcostheta+irsintheta$

Hence, $2 e^{\frac{π}{8} i} = 2 cos (\frac{π}{8}) + 2 sin (\frac{π}{8}) i$ $2e^(pi/8i)=2cos(pi/8)+2sin(pi/8)i$

= $2 \times \frac{\sqrt{2 + \sqrt{2}}}{2} + 2 \times \frac{\sqrt{2 - \sqrt{2}}}{2} i$ $2xxsqrt(2+sqrt2)/2+2xxsqrt(2-sqrt2)/2i$

or in decimal notation approximately

= $2 \times 0.9239 + 2 \times 0.3827 i$ $2xx0.9239+2xx0.3827i$

= $1.8478 + 0.7654 i$ $1.8478+0.7654i$

How can you use trigonometric functions to simplify 2eπ8i 2 e^( ( pi)/8 i ) into a non-exponential complex number?