How can you use trigonometric functions to simplify $2 e^( ( 3 pi)/8 i )$ into a non-exponential complex number?

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Answer 1 · 2018-03-10T17:08:58

$2e^((3pi)/8i)=sqrt(2-sqrt2)+isqrt(2+sqrt2)$

Using trigonometric functions $re^(itheta)$ can be written as

$r(costheta+isintheta)$ or $rcostheta+irsintheta$

Hence $2e^((3pi)/8i)=2(cos((3pi)/8)+isin((3pi)/8))$

= $2(sqrt(2-sqrt2)/2+(sqrt(2+sqrt2)/2)i)$

= $sqrt(2-sqrt2)+isqrt(2+sqrt2)$

See details here for $cos67.5^@$ and $sin67.5^@$

How can you use trigonometric functions to simplify 2 e^( ( 3 pi)/8 i ) 2 e^( ( 3 pi)/8 i ) into a non-exponential complex number?