Prove that all roots of polynomial are located in right half plane, given $f(z) = z^3 -6z^2 + 11z - 6 - a$, $a\in \mathbb C$ with modulus<6. 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:)
 A: 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.
A: From the triangle inequality, for all $z\in\mathbb{C}$ we have
$$ \vert f(z) \vert \geq \vert z-1\vert \vert z-2\vert \vert z-3\vert - \vert a\vert > \vert z-1\vert \vert z-2\vert \vert z-3\vert - 6, \text{ since } \vert a\vert < 6. $$
Now suppose $z$ is a root of the equation $f(z) = 0.$ Then, $\vert f(z) \vert = 0,$ and so $6> \vert z-1\vert \vert z-2\vert \vert z-3\vert.\quad (1)$
Suppose, by way of contradiction, that $\Re(z)<0.$
By writing $z=-x+yi: x,y\in\mathbb{R},\ x>0$ and noting that $-(1+x)<-1,$ we have
$$\vert z-1\vert \vert z-2\vert \vert z-3\vert = \left\vert -(1+x) + yi\right\vert\ \left\vert -(1+x) + yi\right\vert\ \left\vert -(1+x) + yi\right\vert $$
$$ > \vert -1\vert \vert -2\vert \vert -3\vert = 6,\ $$
contradicting $(1).$
