Calculate number of zeros using Rouche theorem

Question

I need to calculate number of zeros of $W(z)=z^5+4z^3+z^2+1$ on: $$1)A=\{z:1<|z|<3\}$$ $$2)B=\{z:\operatorname{Re} z> 0\}$$ So for the first part, let's calculate number of zeros for $|z|<1$ and $|z|<3$. $$I. |z|<1$$ $f(z)=4z^3$ and $g(z)=z^5+z^2+1$, $f,g$ are of course analythic on $|z|<1$ and $|f|>|g|$ on |z|=1. Using Rouche Theorem we know that $f$ and $f+g$ has the same number of zeros. It's easy to check that $f$ has $3$ zeros. $$II. |z|<3$$ Now lets choose for $f(z)=z^5$ and $g(z)=4z^3+z^2+1$. $|f|>|g|$ on |z|=3. From that we conclude that $f+g$ has 5 zeros. Summing up :$W(z)=z^5+4z^3+z^2+1$ has on $A$ $5-3=2$ zeros. I think that I get that part right. $$$$I am uncertain about B. I wanted to use standard method - find the number of zeros on semicircle centered at the origin with radius $R$. My proposition for $f(z)=z^5+1$ and $g(z)=4z^3+z^2$ while for big $R$ it will work for smaller we get $|f|<|g|$.

Answer 1

Part A is correctly answered.

Part B. The answer is $4$.

Take $R>0$ large and try to compute how many roots lie inside the semicircle $|z|=R$ and $\mathrm{Re}\,z>0$. Here we can use the argument principle.

In particular, we shall calculate to total change of the argument of $f(z)$, around the boundary of this semicircle (as $R\to\infty$).

When $|z|=R$, $\,R$ large, the polynomial behaves almost like $z^5$. From $z=-iR$ to $z=iR$, and around the circle counterclockwise, the argument of $z^5$, changes from $-\frac{5\pi}2$ to $\frac{5\pi}2$, i.e, $\frac{5}{2}\cdot 2\pi=5\pi $.

Then from $iR$ to $-iR$, along the imaginary axis, $f(z)=z^5+4z^3+z^2+1$, starts (for $R$ large) with $\pi/2$ (or equivalently $5\pi/2$), and in fact, as $z^2+1=-R^2+1<0$, the argument lies in $(\pi/2, \pi)$ until $z=2i,$ where the argument becomes $\pi$.

Then, when $1<R<2,$ the argument lies in $(\pi, 3\pi/2)$ and at $z=i$ ($R=1$), the argument becomes $3\pi/2$.

In $0<R<1$, the argument lies in $(3\pi/2,2\pi)$, and at $z=0$ it becomes $2\pi$.

In $-1<R<0$, lies in $(2\pi,5\pi/2),$ at $R=-1$ becomes $5\pi/2$.

In $-2<R<-1$ lies in $(5\pi/2, 3/\pi)$, at $z=-2$ becomes $3\pi$.

Finally, at $R<-2$ lies in $(3\pi, 7\pi/2)$ and for $R$ large it is approx. $7\pi/2$. So altogether change $+3\pi$, and adding the previous $+5\pi$ we get $8\pi$ and hence 4 roots in $\{\mathrm{Re}\,z>0\}$.