Question

Number of roots of the equation $z^{10}-z^5-992=0$ with real part negative is

• A. $4$
• B. $3$
• C. $2$
• D. $5$

Answer 1

The correct option is D $5$

$z^{10}-z^5-992=0$

$\Rightarrow$ $(z^5{)}^2-z^5-992=0$

$\Rightarrow$ $(z^5{)}^2-32z^5+31z^5-992=0$

$\Rightarrow$ $z^5(z^5-32)+31(z^5-32)=0$

$\Rightarrow$ $(z^5-32)(z^5+31)=0$

$z^5-32=0$

$z^5=32$

$z=\sqrt[5]{532}=(2{)}^5$

$z^5+31=0$

$z^5=-31$

$z=\sqrt[5]{5-31}=(-1.987{)}^5$

Hence, number of roots of $z$ having negative real parts is $5$

corresponding to $z^5=-31$