I am preparing for my function theory exam and came across this problem.
Let $f(z) := z^8 – 8z^5+2$.
I want to count the number of roots in B:= $\big\{z \ \big| \left| z-2 \right| < \frac{1}{2} \big\}$. From plotting $f$ I know it has one root in $B$.
Using that for $z \in B$ we have $\frac{3}{2} < |z| < \frac{5}{2}$, and trying out several candidates for $g$, I sought upper and lower bounds such that one of the following equations is satisfied on $\partial B$:
If $|f|<|g|$ on $\partial B$, then $\#_{roots} \ (f, B) = \#_{roots} (f+ g, B)$, and
if $|f-g| < |f| + |g|$ on $\partial B$, then $\#_{roots} \ (f, B) = \#_{roots} (g, B)$, while in class we also used
if $|f+g| < |g|$ on $\partial B$, then $\#_{roots} \ (f, B) = \#_{roots} (g, B)$.
Taking a closer look, $g$ has to have one complex root within $B$. I can't even think of a candidate right now.
I don't see a clear path to the solution of my problem.
I believe I have a hard time understanding the argument principle, due to it's various equivalent forms. Practically, do I chose one of its inequalities and stick to it or is it better to chose it depending on the problem I am trying to solve?
I know there are $5$ roots within the unit circle. And I know I can find 3 more roots in $B_2(0)$.
Best Answer
Let $g(z)=(z^3-8)(z^5-\frac14)=z^8-8z^5-\frac14z^3+2$. Then for $|z-2|=\frac12$, which implies $\frac32\le |z|\le\frac52$, $$ |f(z)-g(z)|=\frac14|z|^3\le \frac{5^3}{2^5}\le 4 $$ and $$ |g(z)|=|z-2|\cdot|z^2+2z+4|⋅|z^5-\tfrac14|\ge\frac12⋅(12-|z-2|⋅|z+4|)⋅((\tfrac32)^5-\tfrac14)\ge\frac12⋅8⋅7=28 $$ As these estimates establish $|g(z)|>|g(z)-f(z)|$ for $|z-2|=\frac12$, the polynomials $f$ and $g$ have the same number of roots in the disk $|z-2|<\frac12$ by Rouché's theorem. $g(z)$ has exactly one root, $z=2$, inside that disk.