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How can you use trigonometric functions to simplify # 4 e^( ( 11 pi)/6 i ) # into a non-exponential complex number?

1 Answer

Jul 2, 2016

We get #-2\sqrt{3}+2i#.

Explanation:

Euler's formula #e^{ix}=cos x + i sin x# is the key. Here #x={11\pi}/6#, then:

#e^{i{11\pi}/6}=cos({11\pi}/6)+ i sin({11\pi}/6)#
#=-cos(\pi/6)+i sin(\pi/6)#
#({-\sqrt{3}}/2)+i(1/2)#

Multiplying by 4 then gives the answer.

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