Is there any relationship between Rouché's Theorem and Newton's method

Question

In trying to find complex zeros of equation $3^z+4^z=5^z$, we could divide both sides by $(\sqrt{15})^z$ to transform it into $(\frac{4}{\sqrt{15}})^z=(\sqrt{\frac53})^z-(\sqrt{\frac35})^z$. Let's write $a=\log\sqrt{\frac53}\approx 0.255, b=\log\frac4{\sqrt{15}}\approx 0.032$, we could transform the equation into $f(z)=\sinh(az)-\frac12\exp(bz)=0$.

Let $u_0=-1.785...$ which is real root of $3^x=4^x+5^x$, for any $u<u_0, v>2$ lines $\Re(z)=u, \Re(z)=v, \Im(z)=\frac{(k+\frac12)\pi i}a$ forms some rectangles and it is easy to show that in edges of those rectangle, $|\sinh(az)|\gt |\frac12\exp(bz)|$. By applying Rouché's Theorem, we could get that f(z) has exact one zero inside each rectangle, same as function $\sinh(az)=0$.

Next, starting from each zero of function $\sinh(az)=0$, or $z_0=\frac{k\pi i}a$, and using Newton's method that $z_{h+1}=z_h-\frac{f(z)}{f'(z)}$, computer shows that it will finally converges to the zero of f(z) in the rectangle.

Another very interesting result is that after sorting all zeros of f(z) by imaginary part, the difference of neighbour zeros are very close to an ellipse $\frac{x^2}{3.73^2}+\frac{(y-\frac{\pi}a)^2}{3.96^2}=1$.

Another example is polynomail $z^{n+m}-2z^n+1=0$, we could transform it into $z^m-2+1/z^{-n}=0$ and shows that there're exactly m roots with absolute value more than 1. Similarly, computer shows that starting from any root of $z^m-2=0$, we could reach different root of the equation by using Newton's method.

Is there any special mathematics relationship between the Rouché's Theorem and Newton's method?

Answer 1

We could find that for function $f(z)=z\exp(-2.9z)$ and $g(z)=(\frac13z^2+z+\frac13)\exp(-2.9z)$.

In circle $|z|=1$, we have $|f(z)-g(z)|\le |f(z)|$ But if we started unique root (z=0) of f(z) inside the circle and using Newton's method for g(z), finally, it converges to root of g(z) outside the circle instead of the one inside circle.

It looks like a special exception around z=0:

$-5\le \Re z\le 5, -5\le \Im z\le 5$" />

Points in same color family means they will finally converge to same root and points with darker color converges faster.

$-0.5\le \Re z\le 0.5, -0.5\le \Im z\le 0.5$" />

We could find a small green region around z=0.

Below is corresponding result for $sh(az)=\frac12 exp(bz)$

$sh(az)=\frac12 exp(bz)$ where $-10\le \Re z \le 10, -100\le \Im z\le 100$" />