How can you use trigonometric functions to simplify $6 e^( ( 23 pi)/12 i )$ into a non-exponential complex number?

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Answer 1 · 2016-11-15T13:15:14

$6e^(i(23pi)/12)=(3(sqrt6+sqrt2))/2-i(3(sqrt6-sqrt2))/2$

A complex number $a+ib$ in polar form can be written as $rcostheta+irsintheta$

and using series expansion, it can also be written as $re^(itheta)$

Hence $6e^(i(23pi)/12)=6cos((23pi)/12)+i6sin((23pi)/12)$

= $6cos(2pi-pi/12)+i6sin(2pi-pi/12)$

= $6cos(pi/12)-i6sin(pi/12)$

= $6xx(sqrt6+sqrt2)/4-6i(sqrt6-sqrt2)/4$

= $(3(sqrt6+sqrt2))/2-i(3(sqrt6-sqrt2))/2$

How can you use trigonometric functions to simplify 6 e^( ( 23 pi)/12 i ) 6 e^( ( 23 pi)/12 i ) into a non-exponential complex number?