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IIT study group 10 Online
OpenStudy (anonymous):

Solve?

OpenStudy (anonymous):

|dw:1346419943769:dw|

OpenStudy (anonymous):

@Callisto @satellite73 @eliassaab @Mikael @ghazi @mukushla

OpenStudy (anonymous):

TOO COMPLEX

OpenStudy (anonymous):

Complex number is tooo COMPLEX

OpenStudy (anonymous):

@satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73 @satellite73

OpenStudy (anonymous):

no this is okay really

OpenStudy (anonymous):

you have the the cubed root of 1 there

OpenStudy (anonymous):

K satellite SOLVE IT

OpenStudy (anonymous):

so each time you cube it you get 1

OpenStudy (anonymous):

3 goes in to 333 evenly and therefore leaves a remainder of 1 when you divide in to 334

OpenStudy (anonymous):

so calling \(z=-\frac{1}{2}+\frac{\sqrt{3}}{2}i\) we know \(z^3=1\) and so \(z^{333}=1\) and therefore \(z^{334}=z\)

OpenStudy (anonymous):

answer is \(4+5z\)

OpenStudy (anonymous):

these all are the application of cube root of Unity

OpenStudy (anonymous):

yes i see that

OpenStudy (anonymous):

thxxx

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