Show that the function f(z)=(z^2+1)/(z^3+9) is continuous…
Show that the function $\ f(z)=\dfrac{z^2+1}{z^3+9}\ $ is continuous and bounded in the region $|z|<2$.
Explanation of Method
Locate the poles of $f$ (the cube roots of $-9$), show their modulus exceeds $2$ so none lie in the disk, then bound $|f(z)|$ directly using the triangle and reverse-triangle inequalities on the numerator and denominator.
Step 1 · Locate the poles
$$ z^3+9=0\iff z^3=-9=9e^{i\pi}\iff z_k=9^{1/3}e^{i\pi(2k+1)/3},\quad k=0,1,2 $$
Each pole has modulus $$ |z_k|=9^{1/3}\approx2.08 $$
Step 2 · No poles in the disk
Since $9^{1/3}\approx2.08>2$, all three poles lie strictly outside $|z|<2$. Hence $f$ has no singularities in this region, and being a rational function, is continuous throughout $|z|<2$.
Step 3 · Bound the numerator
For $|z|<2$: $$ |z^2+1|\le|z|^2+1<4+1=5 $$
Step 4 · Bound the denominator from below
By the reverse triangle inequality, for $|z|=r<2$: $$ |z^3+9|\ge\big|9-|z|^3\big|=9-r^3>9-8=1 $$
Step 5 · Combine
$$ \boxed{|f(z)|=\left|\frac{z^2+1}{z^3+9}\right|<\frac5{1}=5\qquad\text{for all }|z|<2} $$
More precisely, $|f(z)|\le\dfrac{r^2+1}{9-r^3}$ for $|z|=r$, an increasing function of $r$ on $[0,2)$ approaching (but never reaching) $5$ as $r\to2^-$, confirming the bound is essentially sharp.
Result
This is the standard pattern for showing a rational function is well-behaved on a disk: locate the (finitely many) poles, confirm they sit outside the region, then convert that distance into an explicit numerical bound via the triangle inequality.