# 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.

EDIT: The polynomial is equivalent with $$(z-1)(z-2)(z-3)=a$$ and it. is easily solved by such property! Hurray:)

• why can't you use conjugation ? Mar 22 at 11:59
• I mean, $z$ being a zero of $f$ does not imply $\bar{z}$ is a zero of $f$, since it only works when $f \in \mathbb{R}[x]$. Mar 22 at 12:01
• @KaviRamaMurthy, I cannot understand what you mean. Our polynomial is equivalent with $(z-1)(z-2)(z-3) = a$, and if $a$ is real, the zero must be positive since $x \leq 0$ implies $(z-1)(z-2)(z-3) \leq -6$. I think the statement is true if $a$ is real with $|a|<6$. Can you please elaborate? Mar 22 at 12:06
• I misread the question and I am deleting my comment. Mar 22 at 12:09
• In general you should include what you know related to the question/possible solution attempts, in the body of the question. Mar 22 at 12:20

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.

• That is so clear and understandable answer! Thank you so much :) Mar 22 at 12:13
• The question defines $f(z) = (z-1)(z-2)(z-3) - a.$ So how can you start with $\vert f(z)\vert = \vert(z-1)(z-2)(z-3) \vert ?$ Mar 22 at 13:29
• I changed the notation. Mar 22 at 13:45
• I still see $|f(z)| =|(z-1)(z-2)(z-3)|$ on my screen. Consider changing your $f(z)$ to $g(z).$ Mar 22 at 14:08

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 $$(1).$$