If $z^3+(3+2i)z+(-1+iy)=0$ ( where $i^2=-1$) has one real root, then the value of $y$ does not belongs to :
- $(2,3)$
- $(-5,-1)$
- $(0,1)$
- $(-2,-1)$
My try: I would like you guys to tell me how to analyze a cubic equation; not just for this particular question but how to analyze a cubic equation in general. Like we have determinant and inequalities related to the determinant to analyze a quadratic equation. How do I analyze a cubic equation for different conditions, like when the equation has one real root, no real root, two equal roots, and all other cases.
An idea: suppose $\;x\;$ is the real root, then:
$$x^3+3x+2xi-1+iy=0\implies\begin{cases}x^3+3x-1=0\\{}\\2x+y=0\end{cases}\implies y=-2x$$
Now, the function $\;x^3+3x-1\;$ is strictly monotone increasing (why?) , and by the MVT it has a root in $\;\left(0,\frac12\right)\;$ which is then its unique real root (again, why?) , and thus $\;y\in(-1,0)\;$ , which means...strangely enough, that all the option given in your question are true.