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  • Help me please, - Complex numbers and quadratic equations - JEE Main-4

z is a complex number satisfying the equation \left ( z-1 \right )=\left ( z^{6}+z^{5}+z^{4}+z^{3}+z^{2}+z+1 \right )=0  if  \alpha _{1},\alpha _{2},\alpha _{3}\cdots ,\alpha _{7}  one least non-negative positive arguments corresponding to solutions, such that \alpha _{1}<\alpha _{2}<\alpha _{3}<\alpha _{4}<\alpha _{5}<\alpha _{6}<\alpha _{7}  then \alpha _{2}+\alpha _{6}  equals

  • Option 1)

    \frac{6\pi }{7}

  • Option 2)

    \frac{12\pi }{7}

  • Option 3)

    \frac{10\pi }{7}

  • Option 4)

    \frac{8\pi }{7}

 

Answers (1)

best_answer

Equation reduces to : Z^{7}-1=0

\Rightarrow Z=\left ( 1 \right )^{\frac{1}{7}}

\Rightarrow Z=\cos \frac{2k\pi }{7}+i\sin \frac{2k\pi }{7}\; where\; k=0,1,2,3,\cdots ,6

\alpha _{1}=0,\alpha _{2}=\frac{2\pi }{7},\alpha _{3}=\frac{4\pi }{7},\alpha _{4}=\frac{6\pi }{7},\alpha _{5}=\frac{8\pi }{7},\alpha _{6}=\frac{10\pi }{7} \; and\; \alpha _{7}=\frac{12\pi }{7}

\therefore \alpha _{2}+\alpha _{6}=\frac{12\pi }{7}

 

nth roots of unity -

z=\left ( 1 \right )^{\frac{1}{n}}\Rightarrow z=\cos \frac{2k\pi }{n}+i\sin \frac{2k\pi }{n}

Where k=0,1,2,......,(n-1)

-

 

  


Option 1)

\frac{6\pi }{7}

This is incorrect

Option 2)

\frac{12\pi }{7}

This is correct

Option 3)

\frac{10\pi }{7}

This is incorrect

Option 4)

\frac{8\pi }{7}

This is incorrect

Posted by

divya.saini

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