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How can you use trigonometric functions to simplify # 12 e^( ( 2 pi)/3 i ) # into a non-exponential complex number?

1 Answer

Apr 29, 2016

#-6+6 sqrt(3)i#

Explanation:

Using the identity #e^(i theta) = cos(theta) + i sin(theta)#
Replacing #(2pi)/3# for #theta#

#e^(((2pi)/3)i)# = #cos((2pi)/3) + i sin((2pi)/3)#

#12 e^(((2pi)/3)i)# = #12(cos((2pi)/3) + i sin((2pi)/3))#

# = 12( -1/2 + i sqrt(3)/2)#

# = -6 + 6sqrt(3)i#

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