Equations and Roots

The equation x^2 = 196 has two solutions: x = 14 and x = -14.

${ z }^{ 10 }-{ z }^{ 5 }+1=0$

Let ${ z }^{ 5 }=w$
$\Rightarrow { w }^{ 2 }-w+1=0$

$\Rightarrow w=\frac { 1\pm \sqrt { 3 }  }{ 2 } =\cos { \frac { \Pi  }{ 3 }  } \pm \sin { \frac { \Pi  }{ 3 }  } =cis\left( \pm \frac { \Pi  }{ 3 }  \right) $


$\Rightarrow { z }^{ 5 }=cis\left( \pm \frac { \Pi  }{ 3 }  \right) $

Case 1:
${ z }^{ 5 }=cis\left( \frac { \Pi  }{ 3 }  \right) $

$\Rightarrow z={ \left( cis\left( \frac { \Pi  }{ 3 }  \right)  \right)  }^{ \frac { 1 }{ 5 }  }=cis\left( \frac { 2k\Pi +\Pi  }{ 15 }  \right) \ $        ...{De Moivre's Theorem}

Where k=0,1,2,3,4.
Therefore number of solutions are 5.

Case 2:
${ z }^{ 5 }=cis\left( -\frac { \Pi  }{ 3 }  \right) $

$\Rightarrow z={ \left( cis\left( -\frac { \Pi  }{ 3 }  \right)  \right)  }^{ \frac { 1 }{ 5 }  }=cis\left( \frac { 2k\Pi -\Pi  }{ 15 }  \right) $       ...{De Moivre's Theorem}

Where k=0,1,2,3,4.
Therefore number of solutions are 5.

From case 1 & case 2 total number of solutions of equation ${ z }^{ 10 }-{ z }^{ 5 }+1=0$ are 10.

Ans: D

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