Determine the number of zeros of the polynomial $f(z)=z^{3}-2z-3$ in the region $A= \{ z : \Re(z) > 0, |\Im(z)| < \Re(z) \}$

Question

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.

Answer 1

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 $A$.

Answer 2

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.