I've asked a similar question to this one, and got a nice answer, but now I am struggling with this one.
Rouché's Theorem: If $f(z)$ and $(g(z)$ are analytic on and inside the contour $C$ and $|f(z)|>|g(z)|$ for all $z$ on C, then $f(z)$ and $f(z)+g(z)$ have the same number of zeros.
I'm asked to show that $h(z)=z^3+8z+23$ has only one zero inside the contour $C_2(0)=\{z:|z-0|=3\}$. Here is some visual evidence that this is true.
Now, I've tried all kinds of choices for $f(z)$ and $g(z)$, but none of them have worked. For example, if I let $f(z)=8z+23$ and $g(z)=z^3$, then I can write
$$|f(z)|=|8z+23|\ge||8z|-|23||=1$$
for all $z$ on the contour $C_3(0)$. However,
$$|g(z)|=|z^3|=|z|^3=27$$
for all $z$ on the contour $C_3(0)$. Thus, I have not shown that $|f(z)|>|g(z)|$ for all $z$ on the contour $C_2(0)$. Here's another image that shows $|f(z)|$ is not greater that $|g(z)|$ for all $z$ on the contour $C_3(0)$.
So, can someone give me an $f(z)$ and a $g(z)$ such that $|f(z)|>|g(z)|$ for all $z$ on the contour $C_3(0)$? And if so, can you share the strategy you used to find them? Thanks.
EDIT: after writing the long proof using Rouché's theorem below, I would like to point out a far easier proof not using Rouché's theorem.
Note that $h(0) = 23$ and $h(-2) = -1$. So $h$ has a zero on $(-2,0)$. Denote it $\beta$. Taking the derivative of $h$ on $\mathbb{R}$, $h'(x) = 3x^2 + 8,$ we see that $h$ has at most one zero on $\mathbb{R}$. As $h$ has real coefficients, the other two roots must be conjugate. Denote them $\alpha$ and $\overline{\alpha}$. The product of the zeros of $h$ is $-23$, so $|\beta \alpha \overline{\alpha}| = |\beta| |\alpha|^2 = 23$ and as $|\beta| <2$, this forces $|\alpha|^2 > 23/2$, hence $|\alpha| > 3$, completing the proof.
Original answer:
The bounding argument we'll need to do later is rather laborious and difficult to just stumble upon by mistake. So in finding this approach it is immensely helpful to know what we're trying to do before we try to make the bounding argument.
This approach is motivated by trying to shift the zero of $f$ on your graph to $-2$. On the picture, this is a very small shift, which hints to us that it might well be possible to make this shift with an application of Rouché's theorem. Another way of seeing the same motivation without explicitly graphing things is that the $z^3 + 8z$ terms in $z^3 + 8z + 23$ are both real and negative on the negative real axis and so directly cancel the $23$ there. So the negative real axis is a nice place to look to modify $z^3 + 8z + 23$ to get a concrete zero, with the intention of factoring out the zero and getting a manageable quadratic.
Trying out points there shows $(-2)^3 + 8(-2) = -24$. So if we can show that $|h(z)| > 1$ for all points $z = 3e^{i\theta}$, we can apply Rouche's theorem and do polynomial long division and to show that $z^3 + 8z + 23$ and $$z^3 + 8z + 24 = (z-2)(z^2 - 2z + 12)$$ have the same number of zeros in the disk of radius $3$. One can either then use the quadratic formula to find the zeros of $z^2 - 2z + 12$ and note that they're not in the region bounded by $C$ or just observe that as $z^2 -2z + 12$ is a real-coefficient polynomial, it has conjugate roots $\alpha$ and $\overline{\alpha}$ with $|\alpha|^2 = 12 > 9$, so that $|\alpha| > 3$. This will complete the proof once we prove our estimate above.
We show $|z^3 + 8z + 23| > 1$ for all $z \in C$ rather directly. This is tedious, but has no real major insights. By looking at real parts, we can put all points where the real part of $z^3 + 8z$ potentially cancels with $23$ problematically to three intervals spread evenly around the circle. On the intervals not containing $-1$, the real part of $8z$ is positive and $z^3$ has negative real part, eliminating concern there.
The interval of concern containing $\theta = \pi$ is given by $(\frac{8}{9}\pi, \frac{10}{9}\pi)$ and is trickier as the real parts do genuinely cancel out to be $0$ at some point there. On this interval, both $\cos(3\theta)$ and $\cos(\theta)$ are negative and $\sin(3\theta), \sin(\theta)$ always have the same sign. We use some first-order estimates on $\cos$ and $\sin$ coming from the fundamental theorem of calculus on the relevant intervals to give us the necessary bounds.
In particular, on $(\frac{8}{9}\pi, \frac{10}{9}\pi)$ one can quickly show $$| \cos(3(\theta-\pi)) | \geq 1 - 9|\theta-\pi|, \quad |\cos(\theta-\pi)| \geq 1 -|\theta-\pi|, \quad |\sin(3(\theta-\pi))| \geq \frac{3}{2}|\theta-\pi|, \quad |\sin(\theta-\pi)| \geq \frac{3}{4} |\theta-\pi|. $$ So we see that on our interval, \begin{align*} \text{Re}(z^3 + 8z) + 23 &\leq - [27(1- 9|\theta-\pi|) + 8(1-|\theta-\pi|)] +23 = -12 + 251 |\theta-\pi|, \\ |\text{Im}(z^3 + 8z + 23)| &\geq 27 (3/2|\theta-\pi|) + 24/4|\theta-\pi| = \left(\frac{81}{2} + 6\right)|\theta-\pi| \geq 46|\theta-\pi|. \end{align*} Our first estimate shows us that the real part of $z^3 + 8z + 23$ for $z = 3e^{i\theta}$ in our problem interval has modulus at least $1$ for $|\theta - \pi| < 11/251$. The second estimate shows us that the imaginary part of $z^3 + 8z + 23$ has modulus at least $1$ for $z = r^{ei\theta}$ with $|\theta - \pi| > 1/46$ in our problem interval. As $1/46 < 11/251,$ we are assured that $|z^3 + 8z + 23| > 1$ on our whole problem interval, and the proof is complete.