How can you use trigonometric functions to simplify # 9 e^( ( 7 pi)/6 i ) # into a non-exponential complex number?

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Answer 1 · 2017-02-01T06:46:35

The answer is #=-9/2(sqrt3+i)#

We need Euler's formula

#e^(itheta)=costheta+isintheta#

Therefore,

#9e^(7/6pi i)=9(cos(7/6pi)+isin(7/6pi))#

#cos(7/6pi)=-sqrt3/2#

#sin(7/6pi)=-1/2#

So,

#9e^(7/6pi i)=9(-sqrt3/2-i/2)#

#=-9/2(sqrt3+i)#