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Question:

a) Determine the number of zeros of the polynomial

$$f(z)=z^{3}-2z-3$$ in the region

$$A= \{ z : Re(z) > 0, |Im(z)| < Re(z) \}$$.

(b) Find the number of zeros of the function

$$g(z) = z^3-2z-3+e^{-z^{2}}$$

in the region A.

Comments: I think this is a fairly difficult problem. I assume that you have to use tools like the principle of the argument and/or Rouché's theorem but I do not know how to do that on this type of a region.

Sid
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  • I made an error when I wrote out the polynomial in (a). I corrected it now. I apologize if someone has started to work on this problem. – Sid Aug 28 '13 at 01:59
  • Also forgot to specify in (b) that we are looking for zeros in the region A. – Sid Aug 28 '13 at 02:05
  • The region $A$ is a wedge with angle $\pi/2$, and it's pretty easy to map that region to the interior of a circle (perhaps going through a half-plane as an intermediate step). So you can change the problem to an equivalent problem where Rouché's theorem is more obviously applicable. – Greg Martin Aug 28 '13 at 02:55
  • That is a very interesting idea. Do you mean by the means of a Möbius transformation? Can you give me the procedure for mapping that region to the interior of a circle? – Sid Aug 28 '13 at 03:35
  • $z\mapsto z^2$, then a Möbius transformation. – Greg Martin Aug 28 '13 at 05:43
  • It's possible to find the zeros in the right half-plane so the Möbius transformation is probably not necessary. However, we will have to change the function $g(z)$ into the function $g(z^{2})$, then try to find the zeros of $g(z^{2})$ in the right half-plane. It is clear that any zeros of $g(z)$ in the wedge will also be the the zeros of $g(z^{2)}$ in the right half-plane, but does this relationship also work the other way? Edit: I doubt there is a one-to-one relationship between the zeros of the functions since g(z^2) should have twice as many zeros as g(z). I wonder how you thought this out. – Sid Aug 28 '13 at 20:35
  • Forgot to tag @GregMartin in my comment. – Sid Aug 28 '13 at 21:16

2 Answers2

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Given part a), as proven by @Lubin, Rouché's theorem can be used to solve b). The boundaries of $A$ can be parametrized as $(1+i)t$, $(1-i)t$, $0\le t<\infty$. Since $f$ is a polynomial, and since $g$ behaves as a polynomial for large $|z|$ we can find $R>0$ such that neither $f$ nor $g$ has any zeros for $|z|>R$. Now let $C$ be the positively oriented simple closed curve given by the boundary of the intersection of $A$ and a closed disc of radius $R+1$ centered at $z=0$. We show that $|f-g|<|f|$ on $C$ and hence $f$ and $g$ will have the same number of zeros within $C$ by Rouché's theorem.

On the part of the boundary where $z(t)=(1+i)t$, $0\le t<\infty$, we have that

$|f-g|=|e^{-z^2}|=|e^{-2it^2}|=1$, since $t\in\mathbb{R}$.

while

$|f|=|z^3-2z-3|=|(1+i)^3t^3-2(1+i)t-3|=\sqrt{(2t^3+2t+3)^2+(2t^3-2t)^2}\ge 3$, since $t\in\mathbb{R}_{\ge 0}$. An almost identical estimate holds on the part where $z(t)=(1-i)t$. The inequality ($|f-g|<|f|$) will obviously hold on the circumference of the circle (if $R$ is chosen big enough). This should suffice to prove that $g$ has one zero in $A$.

Bronski Stenberg
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Part a is easy enough to handle. The polynomial has one real root, since the maximum and minimum points of the (real) graph are both below the $x$-axis. And a glance at a sketch of the graph shows that the one real root is positive. The other two roots are of the form $r_1=\alpha+\beta i$ and $r_2=\alpha-\beta i$. But the sum of the roots is zero, so $\alpha<0$. Thus there’s only one root in the region in question.

Lubin
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  • That seems simple enough, we're in luck that the polynomial in (a) is easy to handle. There is probably no such easy solution to (b) since then we are dealing with a more complicated function. But there is a part of your reasoning which I do not understand: why must the sum of the roots be equal to zero? – Sid Aug 28 '13 at 03:45
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    @Sid, because the coefficient of $z^2$ equals $0$. – njguliyev Aug 28 '13 at 09:30