Isolated Singularities and Laurent Series — Examples

On $0<\mathrm{\mid }z\mathrm{\mid }<1$, $\frac{1}{z(z-1)}=-\frac{1}{z}\cdot \frac{1}{1-z}=-{\sum }_{n\ge 0}{z}^{n-1}\mathrm{.}$

On $\mathrm{\mid }z\mathrm{\mid }>1$, $\frac{1}{z(z-1)}=\frac{1}{z^2}\cdot \frac{1}{1-z^{-1}}={\sum }_{n\ge 0}{z}^{-n-2}\mathrm{.}$

Writing $w=z-1$, on $0<\mathrm{\mid }w\mathrm{\mid }<1$ one has $\frac{1}{z(z-1)}=\frac{1}{w(1+w)}=\frac{1}{w}\cdot \frac{1}{1-(-w)}={\sum }_{n\ge 0}(-1{)}^n{w}^{n-1}\mathrm{.}$

Since the three right-hand sides are different formal Laurent series attached to different annuli, the same rational function really does have different Laurent expansions on different annuli.

For $f(z)=z^m$, the pulled-back function is $f(1\mathrm{/}w)=w^{-m}$ on $0<\mathrm{\mid }w\mathrm{\mid }<1\mathrm{/}R$, so it has a pole of order $m$ at $0$ by [L2]; therefore $f$ has a pole of order $m$ at $\mathrm{\infty }$ by [L1].

For $g(z)=z^{-m}$, the pulled-back function is $g(1\mathrm{/}w)=w^m$, which is holomorphic at $0$ and vanishes there; by [L2] the singularity at $0$ is removable, so $g$ has a removable singularity at $\mathrm{\infty }$ by [L1].

At $0$, the factor $z^{3}f(z)$ is $e^z$, so $f$ has a pole of order $3$ there.

Applying [L1] with $m=3$ gives $\mathrm{Res}\text{\hspace{0.17em}⁣}(\frac{e^z}{z^3},0)=\frac{1}{2!}\frac{d^2}{d{z}^2}(e^z){\mid }_{z=0}\mathrm{.}$

By [L2], the second derivative of $e^z$ is still $e^z$, so the right-hand side of step 2.1 is $\frac{1}{2}{e}^0=\frac{1}{2}$.

At $a=1$ one has $q(1)=0$, $q^{\mathrm{\prime }}(z)=2z$, and therefore $q^{\mathrm{\prime }}(1)=2\mathrm{\ne }0$.

Applying [L1] gives $\mathrm{Res}\text{\hspace{0.17em}⁣}(\frac{z+1}{z^2-1},1)=\frac{p(1)}{q^{\mathrm{\prime }}(1)}=\frac{2}{2}=1.$

The Laurent expansion of $f$ at $0$ is just $z^{-2}$, so the coefficient of $z^{-1}$ is $0$; by [L1], $\mathrm{Res}(f,0)=0$.

The same Laurent expansion has finite nonzero principal part $z^{-2}$, so [L2] makes $0$ a pole of order $2$.

Thus $f$ has residue $0$ at $0$ but the singularity is not removable, refuting the claim.

Since $z\mapsto 1\mathrm{/}z$ is holomorphic on $0<\mathrm{\mid }z\mathrm{\mid }<1$ and $\exp$ is entire by [L2], the function $f(z)=e^{1\mathrm{/}z}$ is holomorphic on $0<\mathrm{\mid }z\mathrm{\mid }<1$.

If $e^{1\mathrm{/}{z}_0}=0$ for some $z_0\mathrm{\ne }0$, then [L3] gives $1=e^{1\mathrm{/}{z}_0}{e}^{-1\mathrm{/}{z}_0}=e^0=0$, impossible. Hence $e^{1\mathrm{/}z}$ never takes the value $0$.

Along the positive real axis, $f(t)=e^{1\mathrm{/}t}\to \mathrm{\infty }$ as $t\downarrow 0$, so the singularity is not removable. Along the negative real axis, $f(-t)=e^{-1\mathrm{/}t}\to 0$ as $t\downarrow 0$, so the singularity is not a pole.

By [L1], a singularity that is neither removable nor a pole is essential, so $0$ is an essential singularity of $e^{1\mathrm{/}z}$.

Thus $e^{1\mathrm{/}z}$ has an essential singularity at $0$ and still omits the value $0$, refuting the claim and showing why Casorati-Weierstrass theorem gives density rather than surjectivity.

For $t>0$, one has $\mathrm{\mid }f(t)\mathrm{\mid }=\mathrm{\mid }\sin (1\mathrm{/}t)\mathrm{\mid }\le 1$, so the function is bounded along the positive real axis approaching $0$.

For $t>0$, $f(it)=\sin (-i\mathrm{/}t)=-i\sinh (1\mathrm{/}t),$ so $\mathrm{\mid }f(it)\mathrm{\mid }=\mathrm{\mid }\sinh (1\mathrm{/}t)\mathrm{\mid }\to \mathrm{\infty }$ as $t\downarrow 0$; therefore the singularity is not removable.

Since $f$ is bounded along the positive real axis by step 1.1, the modulus does not tend to $\mathrm{\infty }$ along every approach to $0$; therefore the singularity is not a pole.

By [L2], a singularity that is neither removable nor a pole is essential, so $0$ is an essential singularity of $\sin (1\mathrm{/}z)$.

For every positive integer $n$, the point $z_n:=1\mathrm{/}(n\pi )$ satisfies $\sin (1\mathrm{/}{z}_n)=\sin (n\pi )=0$ by [L1], so $f$ has a pole at each $z_n$.

The sequence $(z_n)$ tends to $0$, so every punctured neighbourhood of $0$ contains some pole of $f$. Hence no punctured neighbourhood of $0$ is a region on which $f$ is holomorphic, and the singularity at $0$ is not isolated.

By [L1], the function $e^{1\mathrm{/}z}$ has a Laurent expansion on $0<\mathrm{\mid }z\mathrm{\mid }<1$.

If that Laurent expansion had only finitely many negative powers, its principal part would be finite. It cannot be zero, because then the singularity would be removable and hence not essential. So the principal part would be finite and nonzero.

By [L2], a finite nonzero principal part would make the singularity a pole, contradicting [L3]. Therefore the Laurent expansion of $e^{1\mathrm{/}z}$ has infinitely many negative powers.