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:)

  • $\begingroup$ why can't you use conjugation ? $\endgroup$
    – G Cab
    Mar 22 at 11:59
  • $\begingroup$ I mean, $z$ being a zero of $f$ does not imply $\bar{z}$ is a zero of $f$, since it only works when $f \in \mathbb{R}[x]$. $\endgroup$
    – mathhello
    Mar 22 at 12:01
  • 1
    $\begingroup$ @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? $\endgroup$
    – mathhello
    Mar 22 at 12:06
  • $\begingroup$ I misread the question and I am deleting my comment. $\endgroup$ Mar 22 at 12:09
  • 2
    $\begingroup$ In general you should include what you know related to the question/possible solution attempts, in the body of the question. $\endgroup$ Mar 22 at 12:20

2 Answers 2


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.

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


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