Exercises — Residues and poles

What we do: look at the lowest power of $z$ that appears once we expand as a Laurent series — the number of negative powers is the whole story.

(a) $\frac{\sin z}{z}=\frac{z-z^3/6+\dots }{z}=1-\frac{z^2}{6}+\dots$ — no negative powers $\Rightarrow$ removable. (b) $\cos z=1-\frac{z^2}{2}+\dots$, so $\frac{\cos z}{z^3}=z^{-3}-\frac{1}{2}{z}^{-1}+\dots$ — lowest term $z^{-3}$ $\Rightarrow$ ==pole of order $3$==. (c) $e^{1/z}={\sum }_{n\ge 0}\frac{1}{n!}{z}^{-n}$ — infinitely many negative powers $\Rightarrow$ essential. (d) $\frac{z+1}{z^2}=z^{-1}+z^{-2}$ — lowest term $z^{-2}$ $\Rightarrow$ ==pole of order $2$==.

What we do: factor the denominator. $z^2+4=(z-2i)(z+2i)$. Each factor appears to the first power, so $z=2i$ and $z=-2i$ are each simple poles (order $1$). Numerator $1$ never vanishes there, so nothing cancels.

What we do: $z=-3$ is a simple pole (single factor $z+3$). Use $\mathrm{Res}={\lim }_{z\to z_0}(z-z_0)f$. $\mathrm{Res}={\lim }_{z\to -3}(z+3)\cdot \frac{z}{(z-1)(z+3)}=\frac{-3}{(-3-1)}=\frac{-3}{-4}=\frac{3}{4}.$ Why this step: multiplying by $(z+3)$ deletes exactly the offending factor, leaving something finite to substitute.

What we do: $p=e^z$, $q=z^2+9$, $q^{\prime }=2z$. Check: $q(3i)=(3i{)}^2+9=-9+9=0$ ✓, $q^{\prime }(3i)=6i\ne 0$ ✓, $p(3i)=e^{3i}\ne 0$ ✓. $\mathrm{Res}=\frac{p(3i)}{q^{\prime }(3i)}=\frac{e^{3i}}{6i}.$ Why this tool: near a simple root $q(z)\approx q^{\prime }(z_0)(z-z_0)$, so $(z-z_0)f\to p(z_0)/q^{\prime }(z_0)$ — the shortcut just skips the algebra of factoring $q$.

What we do: order-$3$ pole, so use the derivative formula with $m=3$: $\mathrm{Res}=\frac{1}{2!}{\lim }_{z\to 0}\frac{d^2}{d{z}^2}[z^3\cdot \frac{\cos z}{z^3}]=\frac{1}{2}{\lim }_{z\to 0}\frac{d^2}{d{z}^2}\cos z=\frac{1}{2}(-\cos 0)=-\frac{1}{2}.$ Cross-check: from 1.1(b), $a_{-1}=-\frac{1}{2}$ was the $z^{-1}$ coefficient directly. ✓

What we do: two singularities — $z=0$ (order $2$) and $z=1$ (simple). Different tools each.

At $z=1$ (simple): $\mathrm{Res}={\lim }_{z\to 1}(z-1)\frac{1}{z^2(z-1)}=\frac{1}{1^2}=1.$

At $z=0$ (order $2$, $m=2$): $\mathrm{Res}=\frac{1}{1!}{\lim }_{z\to 0}\frac{d}{dz}[z^2\cdot \frac{1}{z^2(z-1)}]={\lim }_{z\to 0}\frac{d}{dz}\frac{1}{z-1}={\lim }_{z\to 0}\frac{-1}{(z-1{)}^2}=-1.$ Sanity check (with the precise condition): define the residue at infinity as the residue of $-\frac{1}{w^2}f(1/w)$ at $w=0$. A general theorem says the sum of all residues — every finite pole plus the point at infinity — is exactly $0$ for any function analytic except at finitely many isolated points. When $f$ is rational and decays faster than $1/z$ as $z\to \infty$ (here $f\sim 1/z^3$, so numerator degree is at least $2$ below denominator degree), the residue at infinity is itself $0$, and the theorem collapses to: the finite residues sum to $0$. Check: $1+(-1)=0$ ✓. (If $f$ only decays like $1/z$, the residue at infinity is generally nonzero and this shortcut does not apply.)

What we do: first the order. Near $0$, $e^z-1=z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots =z(1+\frac{z}{2}+\frac{z^2}{6}+\dots )$. Squaring: $(e^z-1{)}^2=z^2(1+\frac{z}{2}+\dots {)}^2$. The prefactor $z^2$ means an ==order-$2$ pole==.

Residue ($m=2$): let $h(z)=z^{2}g(z)=\frac{z^2}{(e^z-1{)}^2}=(\frac{z}{e^z-1}{)}^2$. First we derive the series of $u(z)=\frac{z}{e^z-1}$ rather than quote it. Write $u=1+c_{1}z+c_2{z}^2+\dots$ and use $e^z-1=z+\frac{z^2}{2}+\frac{z^3}{6}+\dots$. The defining identity is $u\cdot \frac{e^z-1}{z}=1$, i.e. $(1+c_{1}z+c_2{z}^2+\dots )(1+\frac{z}{2}+\frac{z^2}{6}+\dots )=1.$ Match powers of $z$:

• $z^0$: $1=1$ ✓.
• $z^1$: $c_1+\frac{1}{2}=0\Rightarrow c_1=-\frac{1}{2}$.
• $z^2$: $c_2+\frac{1}{2}{c}_1+\frac{1}{6}=0\Rightarrow c_2-\frac{1}{4}+\frac{1}{6}=0\Rightarrow c_2=\frac{1}{12}$.

So $\frac{z}{e^z-1}=1-\frac{z}{2}+\frac{z^2}{12}-\dots$ — derived, not assumed. Squaring (we only need up to $z^1$): $h(z)=(1-\frac{z}{2}+\dots {)}^2=1-z+\dots$ So $\frac{d}{dz}h(0)=$ coefficient of $z$ $=-1$. Hence $\mathrm{Res}=\frac{1}{1!}{h}^{\prime }(0)=-1.$ Why series, not brute derivative: differentiating $z^2/(e^z-1{)}^2$ directly is a $0/0$ nightmare at $z=0$; expanding $z/(e^z-1)$ (analytic, value $1$ at $0$) makes the limit painless.

What we do: find poles, test which lie inside $|z|=3$, sum their residues, multiply by $2\pi i$. Poles at $z=1$ and $z=2$; both satisfy $|z|<3$, so both are enclosed (each with winding number $+1$). ${\mathrm{Res}}_{z=1}={\lim }_{z\to 1}(z-1)f=\frac{1}{(1-2)}=-1,{\mathrm{Res}}_{z=2}={\lim }_{z\to 2}(z-2)f=\frac{2}{(2-1)}=2.$ $\oint =2\pi i(-1+2)=2\pi i.$

What we do: re-test enclosure. Now $|z|=1.5$ contains $z=1$ ($|1|<1.5$ ✓) but not $z=2$ ($|2|>1.5$ ✗). Only $z=1$ counts: ${\mathrm{Res}}_{z=1}=-1$. $\oint =2\pi i(-1)=-2\pi i.$ Punchline: the same function gives a different integral because a different set of whirlpools sits inside the loop — the contour, not the function alone, decides.

What we do: view the real integral as the real-axis piece of a closed contour: the segment $[-R,R]$ along the real axis, plus a big semicircle of radius $R$ arching back through the upper half plane. The figure below shows this exact contour — study it before reading on.

On that semicircle $|f|\sim 1/R^2$ while its length $\sim \pi R$, so the arc integral $\sim \pi R/R^2\to 0$ as $R\to \infty$. What remains is: (real-line integral) $=2\pi i\times$(residues inside the upper half plane). Poles of $\frac{1}{z^2+4}$: $z=\pm 2i$; only $z=2i$ is upper. Quotient shortcut ($q=z^2+4$, $q^{\prime }=2z$): ${\mathrm{Res}}_{z=2i}=\frac{1}{2(2i)}=\frac{1}{4i}.$ ${\int }_{-\infty }^{\infty }\frac{dx}{x^2+4}=2\pi i\cdot \frac{1}{4i}=\frac{\pi }{2}.$ Check: the elementary antiderivative gives $\frac{1}{2}\arctan (x/2){|}_{-\infty }^{\infty }=\frac{1}{2}\pi =\frac{\pi }{2}$ ✓.

What we do: same upper-semicircle setup (same figure as 5.1, just a different pole). Here degree of denominator ($4$) exceeds numerator ($2$) by $2$, so the arc still vanishes. The only upper pole is $z=i$, and it is an order-$2$ pole (because $(z^2+1{)}^2=(z-i{)}^2(z+i{)}^2$). Set $\varphi (z)=(z-i{)}^{2}f(z)=\frac{z^2}{(z+i{)}^2}$. For $m=2$: ${\mathrm{Res}}_{z=i}={\lim }_{z\to i}\frac{d}{dz}\frac{z^2}{(z+i{)}^2}.$ Quotient rule: $\frac{d}{dz}\frac{z^2}{(z+i{)}^2}=\frac{2z(z+i{)}^2-z^2\cdot 2(z+i)}{(z+i{)}^4}=\frac{2z(z+i)-2z^2}{(z+i{)}^3}=\frac{2zi}{(z+i{)}^3}.$ At $z=i$: $(z+i)=2i$, $(2i{)}^3=-8i$, numerator $=2i\cdot i=-2$. So $\mathrm{Res}=\frac{-2}{-8i}=\frac{1}{4i}$. ${\int }_{-\infty }^{\infty }\frac{x^2}{(x^2+1{)}^2}dx=2\pi i\cdot \frac{1}{4i}=\frac{\pi }{2}.$ Why the order matters: treating $z=i$ as simple (multiply by one factor of $(z-i)$) would leave a residual $(z-i)$ in the denominator and blow up — the tell-tale sign you under-counted the order.

What we do: essential singularity — no order-$m$ formula applies, so we expand and pick $a_{-1}$. $z^2{e}^{1/z}=z^2{\sum }_{n=0}^{\infty }\frac{1}{n!}{z}^{-n}={\sum }_{n=0}^{\infty }\frac{1}{n!}{z}^{2-n}.$ The $z^{-1}$ term needs exponent $2-n=-1\Rightarrow n=3$, giving coefficient $\frac{1}{3!}=\frac{1}{6}$. $\mathrm{Res}=\frac{1}{6}.$ Why series only: for an essential singularity $z^{m}f$ never becomes analytic for any finite $m$, so the limit/derivative machinery has nothing to latch onto — the Laurent coefficient is the answer.

• L1 done if you classify singularities from the Laurent series without computing anything.
• L2 done if you pick simple / quotient / order-$m$ correctly and remember $1/(m-1)!$.
• L3 done if you find the true pole order (watch squared denominators) before choosing a tool.
• L4 done if you keep only enclosed poles for each contour.
• L5 done if you justify the vanishing arc and handle essential singularities by series.