Exercises — Residues and poles

Hum kya karte hain: $z$ ki woh sabse chhoti power dekho jo Laurent series mein expand karne par milti hai — negative powers ki count hi poori kahaani hai.

(a) $\frac{\sin z}{z}=\frac{z-z^3/6+\dots }{z}=1-\frac{z^2}{6}+\dots$ — koi negative power nahi $\Rightarrow$ removable. (b) $\cos z=1-\frac{z^2}{2}+\dots$, toh $\frac{\cos z}{z^3}=z^{-3}-\frac{1}{2}{z}^{-1}+\dots$ — sabse chhota term $z^{-3}$ $\Rightarrow$ ==pole of order $3$==. (c) $e^{1/z}={\sum }_{n\ge 0}\frac{1}{n!}{z}^{-n}$ — infinitely saari negative powers $\Rightarrow$ essential. (d) $\frac{z+1}{z^2}=z^{-1}+z^{-2}$ — sabse chhota term $z^{-2}$ $\Rightarrow$ ==pole of order $2$==.

Hum kya karte hain: denominator factor karo. $z^2+4=(z-2i)(z+2i)$. Har factor pehli power mein hai, isliye $z=2i$ aur $z=-2i$ dono simple poles hain (order $1$). Numerator $1$ wahan kabhi vanish nahi karta, toh kuch cancel nahi hota.

Hum kya karte hain: $z=-3$ simple pole hai (single factor $z+3$). $\mathrm{Res}={\lim }_{z\to z_0}(z-z_0)f$ use karo. $\mathrm{Res}={\lim }_{z\to -3}(z+3)\cdot \frac{z}{(z-1)(z+3)}=\frac{-3}{(-3-1)}=\frac{-3}{-4}=\frac{3}{4}.$ Yeh step kyun: $(z+3)$ se multiply karne par exactly woh offending factor delete ho jaata hai, aur kuch finite bachta hai jisme substitute kar sako.

Hum kya karte hain: $p=e^z$, $q=z^2+9$, $q^{\prime }=2z$. Check karo: $q(3i)=(3i{)}^2+9=-9+9=0$ ✓, $q^{\prime }(3i)=6i\ne 0$ ✓, $p(3i)=e^{3i}\ne 0$ ✓. $\mathrm{Res}=\frac{p(3i)}{q^{\prime }(3i)}=\frac{e^{3i}}{6i}.$ Yeh tool kyun: simple root ke paas $q(z)\approx q^{\prime }(z_0)(z-z_0)$, toh $(z-z_0)f\to p(z_0)/q^{\prime }(z_0)$ — shortcut $q$ ko factor karne ka algebra skip kar deta hai.

Hum kya karte hain: order-$3$ pole hai, toh $m=3$ ke saath derivative formula use karo: $\mathrm{Res}=\frac{1}{2!}{\lim }_{z\to 0}\frac{d^2}{d{z}^2}[z^3\cdot \frac{\cos z}{z^3}]=\frac{1}{2}{\lim }_{z\to 0}\frac{d^2}{d{z}^2}\cos z=\frac{1}{2}(-\cos 0)=-\frac{1}{2}.$ Cross-check: 1.1(b) se, $a_{-1}=-\frac{1}{2}$ directly $z^{-1}$ coefficient tha. ✓

Hum kya karte hain: do singularities hain — $z=0$ (order $2$) aur $z=1$ (simple). Dono ke liye alag tools.

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

$z=0$ par (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 (precise condition ke saath): residue at infinity ko define karo $-\frac{1}{w^2}f(1/w)$ ka $w=0$ par residue ke roop mein. Ek general theorem kehta hai ki sab residues ka sum — har finite pole plus infinity wala point — exactly $0$ hai un functions ke liye jo finitely many isolated points ke alawa analytic hain. Jab $f$ rational ho aur $z\to \infty$ par $1/z$ se tez decay kare (yahan $f\sim 1/z^3$, toh numerator degree denominator degree se kam-se-kam $2$ neeche hai), toh infinity par residue khud $0$ hota hai, aur theorem reduce ho jaata hai: finite residues ka sum $0$ hai. Check karo: $1+(-1)=0$ ✓. (Agar $f$ sirf $1/z$ ki tarah decay karta hai, toh infinity par residue generally nonzero hoga aur yeh shortcut nahi lagega.)

Hum kya karte hain: pehle order. $0$ ke paas, $e^z-1=z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots =z(1+\frac{z}{2}+\frac{z^2}{6}+\dots )$. Square karne par: $(e^z-1{)}^2=z^2(1+\frac{z}{2}+\dots {)}^2$. Prefactor $z^2$ ka matlab ==order-$2$ pole== hai.

Residue ($m=2$): maano $h(z)=z^{2}g(z)=\frac{z^2}{(e^z-1{)}^2}=(\frac{z}{e^z-1}{)}^2$. Pehle hum $u(z)=\frac{z}{e^z-1}$ ki series derive karte hain, quote nahi karte. Likho $u=1+c_{1}z+c_2{z}^2+\dots$ aur $e^z-1=z+\frac{z^2}{2}+\frac{z^3}{6}+\dots$ use karo. Defining identity hai $u\cdot \frac{e^z-1}{z}=1$, yaani $(1+c_{1}z+c_2{z}^2+\dots )(1+\frac{z}{2}+\frac{z^2}{6}+\dots )=1.$ $z$ ki powers match karo:

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

Toh $\frac{z}{e^z-1}=1-\frac{z}{2}+\frac{z^2}{12}-\dots$ — derived, assumed nahi. Square karo (sirf $z^1$ tak chahiye): $h(z)=(1-\frac{z}{2}+\dots {)}^2=1-z+\dots$ Toh $\frac{d}{dz}h(0)=$ $z$ ka coefficient $=-1$. Isliye $\mathrm{Res}=\frac{1}{1!}{h}^{\prime }(0)=-1.$ Series kyun, brute derivative kyun nahi: $z^2/(e^z-1{)}^2$ ko directly differentiate karna $z=0$ par $0/0$ ka nightmare hai; $z/(e^z-1)$ (analytic, $0$ par value $1$) expand karna limit ko painless bana deta hai.

Hum kya karte hain: poles dhundo, check karo kaunse $|z|=3$ ke andar hain, unke residues add karo, $2\pi i$ se multiply karo. Poles $z=1$ aur $z=2$ par hain; dono $|z|<3$ satisfy karte hain, toh dono enclosed hain (har ek winding number $+1$ ke saath). ${\mathrm{Res}}_{z=1}={\lim }_{z\to 1}(z-1)f=\frac{1}{(1-2)}=-1,{\mathrm{Res}}_{z=2}={\lim }_{z\to 2}(z-2)f=\frac{2}{(2-1)}=2.$ $\oint =2\pi i(-1+2)=2\pi i.$

Hum kya karte hain: enclosure phir se test karo. Ab $|z|=1.5$ mein $z=1$ hai ($|1|<1.5$ ✓) lekin $z=2$ nahi ($|2|>1.5$ ✗). Sirf $z=1$ count hoga: ${\mathrm{Res}}_{z=1}=-1$. $\oint =2\pi i(-1)=-2\pi i.$ Punchline: wahi function alag integral deta hai kyunki loop ke andar alag set of whirlpools hain — contour decide karta hai, sirf function nahi.

Hum kya karte hain: real integral ko ek closed contour ke real-axis piece ke roop mein dekho: real axis par segment $[-R,R]$, plus radius $R$ ka ek bada semicircle jo upper half plane mein ghoom ke wapas aata hai. Neeche wali figure bilkul yahi contour dikhati hai — aage padhne se pehle use study karo.

Us semicircle par $|f|\sim 1/R^2$ jabki uski length $\sim \pi R$ hai, toh arc integral $\sim \pi R/R^2\to 0$ jab $R\to \infty$. Jo bachta hai: (real-line integral) $=2\pi i\times$(upper half plane ke andar residues). $\frac{1}{z^2+4}$ ke poles: $z=\pm 2i$; sirf $z=2i$ upper mein hai. 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: elementary antiderivative deta hai $\frac{1}{2}\arctan (x/2){|}_{-\infty }^{\infty }=\frac{1}{2}\pi =\frac{\pi }{2}$ ✓.

Hum kya karte hain: wahi upper-semicircle setup (5.1 jaisi same figure, bas alag pole). Yahan denominator ki degree ($4$) numerator ($2$) se $2$ zyada hai, toh arc phir bhi vanish karta hai. Sirf upper pole $z=i$ hai, aur woh order-$2$ pole hai (kyunki $(z^2+1{)}^2=(z-i{)}^2(z+i{)}^2$). Maano $\varphi (z)=(z-i{)}^{2}f(z)=\frac{z^2}{(z+i{)}^2}$. $m=2$ ke liye: ${\mathrm{Res}}_{z=i}={\lim }_{z\to i}\frac{d}{dz}\frac{z^2}{(z+i{)}^2}.$ Quotient rule: $\frac{d}{dz}\frac{z^2}{(z+i{)}^2}=\frac{2z(z+i{)}^2-z^2\cdot 2(z+i)}{(z+i{)}^4}=\frac{2z(z+i)-2z^2}{(z+i{)}^3}=\frac{2zi}{(z+i{)}^3}.$ $z=i$ par: $(z+i)=2i$, $(2i{)}^3=-8i$, numerator $=2i\cdot i=-2$. Toh $\mathrm{Res}=\frac{-2}{-8i}=\frac{1}{4i}$. ${\int }_{-\infty }^{\infty }\frac{x^2}{(x^2+1{)}^2}dx=2\pi i\cdot \frac{1}{4i}=\frac{\pi }{2}.$ Order kyun matter karta hai: $z=i$ ko simple maanna (sirf $(z-i)$ ke ek factor se multiply karna) denominator mein ek residual $(z-i)$ chodh dega aur blow up ho jaayega — yahi tell-tale sign hai ki tumne order undercount kiya.

Hum kya karte hain: essential singularity hai — koi order-$m$ formula apply nahi hota, toh expand karo aur $a_{-1}$ pick karo. $z^2{e}^{1/z}=z^2{\sum }_{n=0}^{\infty }\frac{1}{n!}{z}^{-n}={\sum }_{n=0}^{\infty }\frac{1}{n!}{z}^{2-n}.$ $z^{-1}$ term ke liye exponent chahiye $2-n=-1\Rightarrow n=3$, jo coefficient deta hai $\frac{1}{3!}=\frac{1}{6}$. $\mathrm{Res}=\frac{1}{6}.$ Sirf series kyun: essential singularity ke liye $z^{m}f$ kisi bhi finite $m$ ke liye analytic nahi ban pata, toh limit/derivative machinery ke paas kuch pakadne ko nahi hota — Laurent coefficient hi answer hai.

• L1 done agar tum kuch compute kiye bina Laurent series se singularities classify kar sako.
• L2 done agar tum simple / quotient / order-$m$ sahi se choose karo aur $1/(m-1)!$ yaad rakho.
• L3 done agar tum tool choose karne se pehle sahi pole order dhundo (squared denominators dekho).
• L4 done agar tum har contour ke liye sirf enclosed poles rakho.
• L5 done agar tum vanishing arc justify karo aur essential singularities ko series se handle karo.