Calculate all solutions z ∈ C to 2z^6 + 1 = √3i?

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1 Answer
Shwetank Mauria
Apr 15, 2018

#cos(pi/9)+isin(pi/9)#, #cos((4pi)/9)+isin((4pi)/9)#, #cos((7pi)/9)+isin((7pi)/9)#, #cos((10pi)/9)+isin((10pi)/9)#, #cos((13pi)/9)+isin((13pi)/9)# and #cos((16pi)/9)+isin((16pi)/9)#

Explanation:

As #2z^6+1=sqrt3i#

we have #z^6=-1/2+sqrt3/2i=cos((2pi)/3)+i((2pi)/3)#

Now using De Moivre's theorem

#z=cos(((2pi)/3)/6)+isin(((2pi)/3)/6)=cos(pi/9)+isin(pi/9)#

Further as we also have #z^6=cos(2pi+(2pi)/3)+i(2pi+(2pi)/3)#

and hence roots will repeat after #(2pi)/6=pi/3# and other roots will be

#cos(pi/9+pi/3)+isin(pi/9+pi/3)=cos((4pi)/9)+isin((4pi)/9)#,

#cos((7pi)/9)+isin((7pi)/9)#,

#cos((10pi)/9)+isin((10pi)/9)#,

#cos((13pi)/9)+isin((13pi)/9)# and

#cos((16pi)/9)+isin((16pi)/9)#

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