Answer: Growth Bound Forces Singularity Type

Key Idea / Intuition

The growth rate of $|f(z)|$ near a singularity is the fingerprint of what kind of singularity it is. A pole of order $n$ behaves exactly like $1/|z|^n$ near $0$. An essential singularity has no definite growth rate (it oscillates wildly). A removable singularity stays bounded. Here, $|f(z)| \leq C/|z|^{3/2}$ tells us $f$ blows up no faster than $|z|^{-3/2}$ — but $3/2$ is not an integer! Poles have integer orders. So the singularity cannot be a pole of any integer order. But it also cannot be essential (essential singularities are not bounded by any power of $1/|z|$). The key resolution: multiplying by $z^2$ makes the product bounded near $0$, so the singularity of $z^2 f(z)$ is removable — and working backward pins down exactly what $f$ must look like.

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Formal Proof / Solution

Step 1: Rule out essential singularity.

Near an essential singularity, by Casorati–Weierstrass, $f$ takes values dense in $\mathbb{C}$; in particular it cannot satisfy $|f(z)| \leq C|z|^{-3/2}$ for small $|z|$ (a controlled growth bound). So $z = 0$ is not an essential singularity.

Step 2: Consider $g(z) = z^2 f(z)$.

By hypothesis, $|g(z)| = |z|^2 |f(z)| \leq C |z|^{2 - 3/2} = C |z|^{1/2} \to 0 \quad \text{as } z \to 0.$

So $g(z) \to 0$ as $z \to 0$. Since $g$ is analytic on $0 < |z| < 1$ and bounded near $0$, by Riemann's removable singularity theorem, $g$ extends to an analytic function on the full disk $|z| < 1$ with $g(0) = 0$.

Step 3: Determine $f$.

Since $g(0) = 0$ and $g$ is analytic, we can write $g(z) = z^k h(z)$ for some integer $k \geq 1$ and $h$ analytic with $h(0) \neq 0$.

Therefore $f(z) = \frac{g(z)}{z^2} = z^{k-2} h(z).$

• If $k \geq 2$: then $f(z) = z^{k-2} h(z)$ extends to an analytic function at $0$ (removable singularity).
• If $k = 1$: then $f(z) = z^{-1} h(z)$, a pole of order 1 (simple pole).

Step 4: Check consistency with the growth bound.

• A pole of order $n$ satisfies $|f(z)| \sim C|z|^{-n}$ near $0$.
• The bound $|f(z)| \leq C|z|^{-3/2}$ is satisfied by poles of order $n \leq 3/2$, i.e., $n \leq 1$ (since $n$ must be a non-negative integer).

So $f$ has either:

• A removable singularity at $z = 0$, or
• A pole of order 1 (simple pole) at $z = 0$.

Conclusion:

$\boxed{z = 0 \text{ is either a removable singularity or a simple pole.}}$

The bound $|f(z)| = O(|z|^{-3/2})$ is not tight enough to force a pole, but it rules out poles of order $\geq 2$ and rules out essential singularities entirely. The precise answer: the order of the singularity is at most 1.

Elegant takeaway: The non-integer exponent $3/2$ is the surprise — it forces the singularity to be "between" a simple pole and a removable singularity, and the argument via $z^2 f(z)$ cleanly resolves this.