Question

If $z^3$ + (3 + 2i)z + (-1 + ia) = 0 has one real root , then value lies in interval of

• A. (-z , -1)
• B. (-1, 0)
• C. (0, 1)
• D. (1, 2)

Answer 1

Answer: (B)

(-1, 0)

$z^3+(3+2i)z+(-1+ia)=0$

Say $z=\alpha$ is the real root of the above equation.Hence substituting

$\Rightarrow {\alpha }^3+(3+2i)\alpha +(-1+2a)=0\Rightarrow ({\alpha }^3+3\alpha -1)+i(a+2\alpha )=0$

equating we get

${\alpha }^3+3\alpha -1=0a+2\alpha =0\Rightarrow \alpha =\frac{-a}{2}$

Putting $\alpha =\frac{-a}{2}$ in ${\alpha }^3+3\alpha -1=0$ we get

${\left(\frac{-a}{2}\right)}^3+3\left(\frac{-a}{2}\right)-1=0\Rightarrow \frac{-a^3}{8}-\frac{3a}{2}-1=0\Rightarrow a^3+12a+8=0$

Consider the function $f\left(a\right)=a^3+12a+8$

Now let's check for the sign of the function at points $(-2,1,0,1,3)$

$f\left(0\right)=8>0f(-1)=-1-12+8=-5<0f(-2)=-8-24+8<0f\left(1\right)=1+12+8>0f\left(3\right)=27+36+8>0$