Assuming that \(| b | < 1\) , show that \(f ( z ) = z ^ { 3 } + 3 z ^ { 2 } + b z + b ^ { 2 }\) has exactly two roots (counting multiplicity) in \(| z | < 1\)
Solution
1. On the unit circle,\[ \abs{z^3+bz+b^2}<3. \]
If \(\abs{z}=1\), then \[ \begin{aligned} \abs{z^3+bz+b^2} &\leq \abs{z}^3 +\abs{b}\abs{z} +\abs{b}^2\\ &= 1+\abs{b}+\abs{b}^2. \end{aligned} \] Since \(\abs{b}<1\), \[ 1+\abs{b}+\abs{b}^2<3. \]
2. On the unit circle,\[ \abs{z^3+bz+b^2} < \abs{3z^2}. \]
By step 1, the left-hand side is less than \(3\). If \(\abs{z}=1\), then \[ \abs{3z^2}=3. \]
3. The polynomial\[ f(z)=z^3+3z^2+bz+b^2 \]has exactly two zeros in \(\abs{z}<1\), counting multiplicity.
Write \[ f(z)=3z^2+(z^3+bz+b^2). \] Step 2 is the strict Rouché inequality on \(\abs{z}=1\). Hence \(f\) and \(3z^2\) have the same number of zeros in the unit disk, counting multiplicity. The polynomial \(3z^2\) has exactly two zeros there, both at \(z=0\) counted with multiplicity.
4. Q.E.D.
Step 3 is the required conclusion.