Exercises — Laurent series — principal part, annulus of convergence

(a) Principal part sirf negative-power terms ka sum hai: $\frac{3}{z^2}-\frac{5}{z}.$ (b) Sabse negative power $z^{-2}$ hai aur uska coefficient $3\ne 0$ hai, isliye principal part $(z-0{)}^{-2}$ par ruk jaata hai → pole of order 2. (c) Residue $=c_{-1}=$ coefficient of $\frac{1}{z}=\boxed{-5}$.

(i) Infinitely many negative powers → essential singularity. (ii) Koi bhi negative power nahi → principal part zero hai → removable singularity (function wahan secretly analytic hai).

Us Taylor series se shuru karo jis par hum already bharosa karte hain (Taylor series): $\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots$ Ye step kyun? Numerator analytic hai — $z^3$ se divide karna bas har power ko $3$ neeche shift kar deta hai, aur theek isi se negative powers aati hain. $\frac{\sin z}{z^3}=\frac{1}{z^2}-\frac{1}{3!}+\frac{z^2}{5!}-\cdots$

• Principal part: $\frac{1}{z^2}$ (the $z^{-1}$ term absent hai — uska coefficient $0$ hai).
• Sabse negative power $z^{-2}$ hai coefficient $1\ne 0$ ke saath → pole of order 2.
• Residue $=c_{-1}=\boxed{0}$. $0<|z|<\infty$ mein valid.

$e^z=1+z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots$ $\frac{e^z}{z}=\frac{1}{z}+1+\frac{z}{2}+\frac{z^2}{6}+\cdots$ Pehle chaar 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) $|z|<3$ — yahan $|z|$ chhota hai, isliye badi quantity factor out karo, jo $3$ hai: $\frac{1}{z-3}=\frac{-1}{3}\cdot \frac{1}{1-z/3}=-\frac{1}{3}{\sum }_{n=0}^{\infty }\frac{z^n}{3^n}=-{\sum }_{n=0}^{\infty }\frac{z^n}{3^{n+1}}.$ Valid kyunki $|z/3|<1$. Koi principal part nahi ($3$ par singularity is disk ke bahar hai), isliye yeh ek plain Taylor series hai. $0$ par residue $0$ hai (koi $z^{-1}$ nahi).

(b) $|z|>3$ — ab $|z|$ bada hai, isliye $z$ factor out karo (badi 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 kyunki $|3/z|<1$. Saari negative powers. Residue $=c_{-1}=3^0=\boxed{1}$. (Upar figure mein "large $z$" arrow dekho — bada $|z|$ hamesha tumhe $1/z$ ki taraf dhakelta hai.)

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

Ring $1<|z|<3$ mein: $1$ par singularity andar hai, $3$ wali bahar hai.

• $\frac{1}{z-1}$: kyunki $|z|>1$, $z$ factor out karo → negative powers: $\frac{1}{z-1}=\frac{1}{z}{\sum }_{n\ge 0}\frac{1}{z^n}={\sum }_{n\ge 0}\frac{1}{z^{n+1}}.$
• $\frac{1}{z-3}$: kyunki $|z|<3$, $3$ factor out karo → positive powers: $\frac{1}{z-3}=-{\sum }_{n\ge 0}\frac{z^n}{3^{n+1}}.$ $\pm \frac{1}{2}$ ke saath combine karo: $f(z)=-\frac{1}{2}{\sum }_{n\ge 0}\frac{1}{z^{n+1}}-\frac{1}{2}{\sum }_{n\ge 0}\frac{z^n}{3^{n+1}}.$ $0$ par Residue = $z^{-1}$ ka coefficient = $-\frac{1}{2}$ (pehle sum ke $n=0$ se) $=\boxed{-\frac{1}{2}}$.

Dono singularities ab andar hain, isliye dono pieces $1/z$ mein jaate hain. $\frac{1}{z-1}={\sum }_{n\ge 0}\frac{1}{z^{n+1}},\frac{1}{z-3}=\frac{1}{z}{\sum }_{n\ge 0}\frac{3^n}{z^n}={\sum }_{n\ge 0}\frac{3^n}{z^{n+1}}.$ Coefficients $-\frac{1}{2}$ aur $+\frac{1}{2}$ ke saath: $f(z)=\frac{1}{2}{\sum }_{n\ge 0}\frac{3^n-1}{z^{n+1}}.$ Saari negative powers. $0$ par Residue = $z^{-1}$ ka coefficient = $\frac{1}{2}(3^0-1)=\frac{1}{2}(1-1)=\boxed{0}$. Sanity note: $f$ ki $0$ par koi singularity nahi hai, isliye "residue at $0$" ka $0$ hona consistent hai — outer-region expansion phir bhi $z^{-1}$ term list karta hai, lekin uska coefficient $0$ ho jaata hai.

Use karo $e^{1/z}={\sum }_{n\ge 0}\frac{1}{n!z^n}=1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\cdots$ Har term ko $z^2$ se multiply karo (powers upar $2$ shift ho jaate hain): $z^2{e}^{1/z}=z^2+z+\frac{1}{2!}+\frac{1}{3!z}+\frac{1}{4!z^2}+\cdots$ Phir bhi infinitely many negative powers hain → essential singularity ($z^2$ se multiply karna ek infinite tail ko khatam nahi kar sakta). Residue $=c_{-1}=\frac{1}{3!}=\boxed{\frac{1}{6}}$.

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

• Sabse negative power $z^{-2}$, coefficient $-1\ne 0$ → pole of order 2.
• Residue $=c_{-1}=\boxed{-1}$.
• Residue theorem se, loop $|z|=\frac{1}{2}$ ke saath jo sirf $0$ par singularity enclose karta hai ($1$ par wali bahar hai): ${\oint }_{|z|=1/2}fdz=2\pi i(c_{-1})=2\pi i(-1)=\boxed{-2\pi i}.$ Isliye residues contour integration run karte hain: sirf $z^{-1}$ term loop integral mein survive karta hai (parent se "$k=-1$ gives $2\pi i$" wala fact).

Residue theorem kehta hai closed loop integral $=2\pi i\times$ (sum of residues enclosed).

• $|z|=3$ $z=2$ ko enclose karta hai (kyunki $|2|=2<3$). Sirf wahi pole contribute karta hai: ${\oint }_{|z|=3}gdz=2\pi i(4)=\boxed{8\pi i}.$
• $|z|=1.5$ $z=2$ ko enclose nahi karta (kyunki $2>1.5$), aur $g$ andar baaki jagah analytic hai: ${\oint }_{|z|=1.5}gdz=2\pi i(0)=\boxed{0}.$ Kyun hum $g$ jaane bina answer de sakte hain: loop integral sirf uske andar wali singularities ki parwah karta hai (pole structure aur $k=-1$ survival fact ka consequence).

Principal part mein $-\frac{1}{z^2}$ hai, jiska exponent $-2$ non-zero coefficient ke saath hai. Isliye pole order $2$ hai, $1$ nahi — claim inconsistent hai. Sahi classification hai pole of order 2. Residue $z^{-1}$ ka coefficient hai, jo $\boxed{2}$ hai.

Laurent series ka kaun sa term pole order decide karta hai? ::: Sabse negative power jiska non-zero coefficient ho. Kaun sa term residue decide karta hai? ::: $(z-a{)}^{-1}$ ka coefficient $c_{-1}$. $|z|>3$ mein, kya tum $\frac{1}{z-3}$ ko $z$ ki powers mein expand karoge ya $1/z$ ki powers mein? ::: $1/z$ ki powers mein ($z$ factor out karo, kyunki $|3/z|<1$). Kya $z^2$ se multiply karna essential singularity ko pole mein badal deta hai? ::: Nahi — infinite negative tail kisi bhi finite power shift mein survive kar leti hai. ${\oint }_{|z|=1/2}\frac{dz}{z^2(z-1)}$? ::: $-2\pi i$ (residue at $0$ is $-1$).