Let $f(z) = z^3 -6z^2 + 11z – 6 – a$, where $a$ is a complex number with the modulus less than 6. What I was attempting to show is that all zeroes of $f(z)$ have a positive real part, i.e. located in right half plane.
I searched for results in complex analysis like the Routh-Hurwitz theorem, Gauss-Lucas Theorem, and Rouche theorem, but I have not learned complex analysis so I do not really know how to apply those theorems. Plus, this is a Precalculus test problem. I assume that there exists an elementary proof.
My partial results are that there are neither imaginary roots nor negative real roots, which are pretty obvious….. I am stuck because $f$ is complex-coefficient polynomial and I cannot use properties like conjugate symmetry.
Thanks in advance!
EDIT: The polynomial is equivalent with $(z-1)(z-2)(z-3)=a$ and it. is easily solved by such property! Hurray:)
Best Answer
If the real part $x$ of $z = x +iy$ is $< 0$,
$$|f(z)| =|(z-1)(z-2)(z-3)| = |z-1||z-2||z-3| = \sqrt{(x-1)^2 + y^2}\sqrt{(x-2)^2 + y^2}\sqrt{(x-3)^2 + y^2}\geq (1-x)(2-x)(3-x) > 6$$
Thus, $f(z)$ cannot be equal to $a$, which has modulus less than 6.