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#