If the number of real roots of $x^9-x^5+x^4-1=0$ is $n$, the number of complex roots having argument on…
If the number of real roots of $x^9-x^5+x^4-1=0$ is $n$, the number of complex roots having argument on imaginary axis is $m$ and the number of complex roots having argument in $2^{\text {nd }}$ quadrant is k , then $m \cdot n \cdot k \cdot=$
6
9
12
24
Solution
$\begin{aligned}
& x^9-x^5+x^4-1=0 \Rightarrow\left(x^5+1\right)\left(x^4-1\right)=0 \\
& \Rightarrow x^4=1 \Rightarrow x=1,-1, i,-i \\
& \Rightarrow x^5=-1 \Rightarrow x=-1, e^{-\frac{i \pi}{5}}, e^{\frac{i \pi}{5}}, e^{-\frac{3 \pi i}{5}}, e^{\frac{i 3 \pi}{5}}
\end{aligned}$ No. of real roots is $n=3, m=2, k=1$
$\therefore m \cdot n \cdot k=3 \times 2 \times 1=6$