Question

The number of solutions of equation $z^{10}-z^5+1=0$ are

• A. only two solution
• B. No solution
• C. only five solution
• D. exactly 10

Answer 1

Answer: (D)

exactly 10

$z^{10}-z^5+1=0$
Let $z^5=w$
$\Rightarrow w^2-w+1=0$
$\Rightarrow w=\frac{1\pm \sqrt{3}}{2}=\cos \frac{\Pi }{3}\pm \sin \frac{\Pi }{3}=cis(\pm \frac{\Pi }{3})$
$\Rightarrow z^5=cis(\pm \frac{\Pi }{3})$
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:
$z^5=cis(-\frac{\Pi }{3})$
$\Rightarrow z={\left(cis(-\frac{\Pi }{3})\right)}^{\frac{1}{5}}=cis\left(\frac{2k\Pi -\Pi }{15}\right)$ ...{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