Let $B_r$ denote the closed disk $\{z \in \Bbb C: |z| \le r\}$. State whether $\infty$ is a removable singularity (RS), pole (P), or
essential singularity (ES) in each of the following cases. There may be more than one possibility in each
case.

(a) $f$ is a non-constant polynomial in $z$.

(b) $f(z) =\frac{p(z)}{q(z)}$, where $p,q$ are non-zero polynomials of the same degree.

(c) f is an entire function for which $f^{-1}(B_1)$ is bounded.

(d) f is an entire function for which $f^{-1}(B_r)$ is bounded for all $r > 0$.

Let $f$ be a complex function, then its Laurent series about the point $c$ is given by $$f(z) = \sum\limits_{n=-\infty}^{\infty}a_n(z-c)^n.$$

$c$ is a removable singularity of $f$ if and only if the above series has no terms with negative powers of $z$.

$c$ is a pole of $f$ if and only if the above series has finitely many terms with negative powers of $z$.

$c$ is an essential singularity of $f$ if and only if the above series has infinitely many negative terms.

Now, we shall answer the above question.

Suppose $\infty$ is a removable singularity for the function given in the problem, then by this result $f$ has to be constant, say $f(z) = \alpha$, for some $\alpha \in \Bbb C$. There exists $r>0$ such that $B_r$ contains $\alpha$ and we have $f^{-1}(B_1) = \Bbb C$ which is not bounded. Contradiction. So

(a) $f$ is a non-constant polynomial in $z$.

(b) $f(z) =\frac{p(z)}{q(z)}$, where $p,q$ are non-zero polynomials of the same degree.

(c) f is an entire function for which $f^{-1}(B_1)$ is bounded.

(d) f is an entire function for which $f^{-1}(B_r)$ is bounded for all $r > 0$.

**Solution**:Let $f$ be a complex function, then its Laurent series about the point $c$ is given by $$f(z) = \sum\limits_{n=-\infty}^{\infty}a_n(z-c)^n.$$

**Result**:$c$ is a removable singularity of $f$ if and only if the above series has no terms with negative powers of $z$.

$c$ is a pole of $f$ if and only if the above series has finitely many terms with negative powers of $z$.

$c$ is an essential singularity of $f$ if and only if the above series has infinitely many negative terms.

**Definition**: $\infty$ is a removable singularity (resp, pole or essential singularity) of $f(z)$ if and only if $0$ is a removable singularity (resp, pole or essential singularity) of $f(\frac{1}{z})$.Now, we shall answer the above question.

**option a**: $f(z)$ is a non-constant polynomial in $z$, then clearly $f(\frac{1}{z})$ has finitely many negative terms. Now, by the above result, $0$ is a pole of $f(\frac{1}{z})$ and hence $\infty$ is a pole of $f(z)$ by the definition.**option b**: Given that $f(z) = \frac{p(z)}{q(z)}$ where $p$ and $q$ are non-constant polynomials of same degree (say n). We observe that, after simplification $f(\frac{1}{z}) = \frac{p_1(z)}{q_1(z)}$ where $p_1$ and $q_1$ are again polynomials of same degree. This shows that $\lim\limits{z \to 0}f(\frac{1}{z}) = \frac{\text{constant terms of } p_1}{\text{constant terms of } q_1}$ which is finite and hence removable singularity.**option c**: Let $f$ be an entire function such that $f^{-1}(B_1)$ is bounded.**Result**: Image of any neighborhood of an essential singularity is dense in $\Bbb C$.**Claim**: $\infty$ is**not an essential singularity**of $f$. Suppose $\infty$ is an essential singularity of $f$. We derive a contradiction. Let $B$ be a closed and bounded disc containing $f^{-1}(B_1)$, this is possible since it is a bounded set. Consider $D = \Bbb C \backslash B$, then $D$ is a neighborhood of infinity. Now, by the above result, we have $f(D)$ is dense in the codomain $\Bbb C$. This means that $f(D)$ intersects all the open discs in $\Bbb C$ and, in particular, it should intersect the open unit disc about zero in the co-domain $\Bbb C$. Let $z_0$ be a point in this disc which is in $f(D)$, then $f^{-1}(z_0) \in D = \Bbb C \backslash B$ and $f^{-1}(z_0) \in B$ since $z_0 \in B_1$. Contradiction.**Claim**:**Removable singularity and pole are possible**. Consider the functions $f(z) = 0$ and $g(z)=z$, then both the functions satisfies the given condition. We have $f(\frac{1}{z}) = 0$ and this has no terms of negative powers of $z$ and hence $0$ is a removable singularity of $f(\frac{1}{z})$. In turn, $\infty$ is a**removable singularity**of $f$. Now $g(\frac{1}{z}) = \frac{1}{z}$ which has finitely many negative powers of $z$. So $\infty$ is a**pole**of $g(z)$.**option d**: Similar to option (c). $\infty$ is an**essential singularity is not possible**by the same argument.**Result**: $\infty$ is a removal singularity of $f(z)$ if and only if $f$ is a constant polynomialSuppose $\infty$ is a removable singularity for the function given in the problem, then by this result $f$ has to be constant, say $f(z) = \alpha$, for some $\alpha \in \Bbb C$. There exists $r>0$ such that $B_r$ contains $\alpha$ and we have $f^{-1}(B_1) = \Bbb C$ which is not bounded. Contradiction. So

**removable singularity not possible**. Consider $f(z)=z$, then it satisfies all the given condition of option (d) and has $\infty$ as a**pole**.**Share to your groups:**
In option c ,I think u should consider f(z)=some constant greater than 1 otherwise as u consider 0€B1 and f^-1(B1)=C which is unbounded contradicts

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