Exercises — Laurent series — principal part, annulus of convergence

(a) The principal part is the sum of negative-power terms only: $\frac{3}{z^2}-\frac{5}{z}.$ (b) The most negative power is $z^{-2}$ and its coefficient $3\ne 0$, so the principal part stops at $(z-0{)}^{-2}$ → pole of order 2. (c) Residue $=c_{-1}=$ coefficient of $\frac{1}{z}=\boxed{-5}$.

(i) Infinitely many negative powers → essential singularity. (ii) No negative powers at all → principal part is zero → removable singularity (the function is secretly analytic there).

Start from the Taylor series we already trust (Taylor series): $\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots$ Why this step? The numerator is analytic — dividing by $z^3$ just shifts every power down by $3$, which is exactly how the negative powers appear. $\frac{\sin z}{z^3}=\frac{1}{z^2}-\frac{1}{3!}+\frac{z^2}{5!}-\cdots$

• Principal part: $\frac{1}{z^2}$ (the $z^{-1}$ term is absent — its coefficient is $0$).
• Most negative power is $z^{-2}$ with coefficient $1\ne 0$ → pole of order 2.
• Residue $=c_{-1}=\boxed{0}$. Valid in $0<|z|<\infty$.

$e^z=1+z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots$ $\frac{e^z}{z}=\frac{1}{z}+1+\frac{z}{2}+\frac{z^2}{6}+\cdots$ First four terms: $\frac{1}{z}+1+\frac{z}{2}+\frac{z^2}{6}$. Residue $=c_{-1}=\boxed{1}$. Pole of order 1 (a simple pole).

(a) $|z|<3$ — here $|z|$ is small, so factor out the bigger quantity, which is $3$: $\frac{1}{z-3}=\frac{-1}{3}\cdot \frac{1}{1-z/3}=-\frac{1}{3}{\sum }_{n=0}^{\infty }\frac{z^n}{3^n}=-{\sum }_{n=0}^{\infty }\frac{z^n}{3^{n+1}}.$ Valid because $|z/3|<1$. No principal part (singularity at $3$ sits outside this disk), so this is a plain Taylor series. Residue at $0$ is $0$ (no $z^{-1}$).

(b) $|z|>3$ — now $|z|$ is big, so factor out $z$ (the bigger quantity): $\frac{1}{z-3}=\frac{1}{z}\cdot \frac{1}{1-3/z}=\frac{1}{z}{\sum }_{n=0}^{\infty }\frac{3^n}{z^n}={\sum }_{n=0}^{\infty }\frac{3^n}{z^{n+1}}.$ Valid because $|3/z|<1$. All negative powers. Residue $=c_{-1}=3^0=\boxed{1}$. (See the "large $z$" arrow in the figure above — big $|z|$ always pushes you toward $1/z$.)

Partial fractions: $\frac{1}{(z-1)(z-3)}=\frac{-\frac{1}{2}}{z-1}+\frac{\frac{1}{2}}{z-3}$. Check: $\frac{1}{2}(\frac{1}{z-3}-\frac{1}{z-1})=\frac{1}{2}\cdot \frac{(z-1)-(z-3)}{(z-1)(z-3)}=\frac{1}{2}\cdot \frac{2}{(z-1)(z-3)}$. ✔

In the ring $1<|z|<3$: the singularity at $1$ is inside, the one at $3$ is outside.

• $\frac{1}{z-1}$: since $|z|>1$, factor out $z$ → negative powers: $\frac{1}{z-1}=\frac{1}{z}{\sum }_{n\ge 0}\frac{1}{z^n}={\sum }_{n\ge 0}\frac{1}{z^{n+1}}.$
• $\frac{1}{z-3}$: since $|z|<3$, factor out $3$ → positive powers: $\frac{1}{z-3}=-{\sum }_{n\ge 0}\frac{z^n}{3^{n+1}}.$ Combine with the $\pm \frac{1}{2}$: $f(z)=-\frac{1}{2}{\sum }_{n\ge 0}\frac{1}{z^{n+1}}-\frac{1}{2}{\sum }_{n\ge 0}\frac{z^n}{3^{n+1}}.$ Residue at $0$ = coefficient of $z^{-1}$ = $-\frac{1}{2}$ (from $n=0$ of the first sum) $=\boxed{-\frac{1}{2}}$.

Both singularities are now inside, so both pieces go into $1/z$. $\frac{1}{z-1}={\sum }_{n\ge 0}\frac{1}{z^{n+1}},\frac{1}{z-3}=\frac{1}{z}{\sum }_{n\ge 0}\frac{3^n}{z^n}={\sum }_{n\ge 0}\frac{3^n}{z^{n+1}}.$ With coefficients $-\frac{1}{2}$ and $+\frac{1}{2}$: $f(z)=\frac{1}{2}{\sum }_{n\ge 0}\frac{3^n-1}{z^{n+1}}.$ All negative powers. Residue at $0$ = coefficient of $z^{-1}$ = $\frac{1}{2}(3^0-1)=\frac{1}{2}(1-1)=\boxed{0}$. Sanity note: $f$ has no singularity at $0$, so the "residue at $0$" being $0$ is consistent — the outer-region expansion still lists a $z^{-1}$ term, but its coefficient collapses to $0$.

Use $e^{1/z}={\sum }_{n\ge 0}\frac{1}{n!z^n}=1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\cdots$ Multiply every term by $z^2$ (shifts powers up by 2): $z^2{e}^{1/z}=z^2+z+\frac{1}{2!}+\frac{1}{3!z}+\frac{1}{4!z^2}+\cdots$ There are still infinitely many negative powers → essential singularity (multiplying by $z^2$ cannot kill an infinite tail). Residue $=c_{-1}=\frac{1}{3!}=\boxed{\frac{1}{6}}$.

In $|z|<1$ expand $\frac{1}{z-1}=-\frac{1}{1-z}=-{\sum }_{n\ge 0}{z}^n=-(1+z+z^2+\cdots )$. Then $\frac{1}{z^2(z-1)}=\frac{1}{z^2}(-(1+z+z^2+\cdots ))=-\frac{1}{z^2}-\frac{1}{z}-1-z-\cdots$

• Most negative power $z^{-2}$, coefficient $-1\ne 0$ → pole of order 2.
• Residue $=c_{-1}=\boxed{-1}$.
• By the Residue theorem, with the loop $|z|=\frac{1}{2}$ enclosing only the singularity at $0$ (the one at $1$ is outside): ${\oint }_{|z|=1/2}fdz=2\pi i(c_{-1})=2\pi i(-1)=\boxed{-2\pi i}.$ This is exactly why residues run contour integration: only the $z^{-1}$ term survives the loop integral (the "$k=-1$ gives $2\pi i$" fact from the parent).

The residue theorem says a closed loop integral $=2\pi i\times$ (sum of residues enclosed).

• $|z|=3$ encloses $z=2$ (since $|2|=2<3$). Only that pole contributes: ${\oint }_{|z|=3}gdz=2\pi i(4)=\boxed{8\pi i}.$
• $|z|=1.5$ does not enclose $z=2$ (since $2>1.5$), and $g$ is analytic everywhere else inside: ${\oint }_{|z|=1.5}gdz=2\pi i(0)=\boxed{0}.$ Why we can answer without knowing $g$: the loop integral cares only about singularities inside it (a consequence of the pole structure plus the $k=-1$ survival fact).

The principal part contains $-\frac{1}{z^2}$, whose exponent $-2$ has a non-zero coefficient. So the pole order is $2$, not $1$ — the claim is inconsistent. The correct classification is pole of order 2. The residue is the coefficient of $z^{-1}$, which is $\boxed{2}$.

Which term of a Laurent series decides the pole order? ::: The most negative power with a non-zero coefficient. Which term decides the residue? ::: The coefficient $c_{-1}$ of $(z-a{)}^{-1}$. In $|z|>3$, do you expand $\frac{1}{z-3}$ in powers of $z$ or of $1/z$? ::: In powers of $1/z$ (factor out $z$, since $|3/z|<1$). Does multiplying by $z^2$ turn an essential singularity into a pole? ::: No — an infinite negative tail survives any finite power shift. ${\oint }_{|z|=1/2}\frac{dz}{z^2(z-1)}$? ::: $-2\pi i$ (residue at $0$ is $-1$).