Question

Number of roots of the equation $z^{10}-z^5-992=0$ where real parts are negative is

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (C)

$5$

We have,
$z^{10}-z^5-992=0$
Put $z^5=u$
$\Rightarrow u^2-u-992=0$
$\Rightarrow u=\frac{1\pm \sqrt{1+4\times 992}}{2}$
$=\frac{1\pm 63}{2}=32,-31$
$\Rightarrow z^5=32,(-1{)}^{5}31$
$\Rightarrow z=(32{)}^{1/5},-(31{)}^5$
Hence number of roots of $z$ having negative real parts is $5$
corresponding to $z^5=-31$