Question bank — Laurent series — principal part, annulus of convergence

False — agar $f$ poori disk par analytic hai toh koi inner singularity nahi hai, isliye har $c_{-n}=0$ aur Laurent series collapse hokar ek ordinary Taylor series ban jaati hai.

True — jab principal part zero hoti hai toh Laurent series hi Taylor series hoti hai; Taylor, Laurent ka special case hai jisme koi negative powers nahi hote.

False — residue $c_{-1}$ hota hai, yaani $(z-a{)}^{-1}$ ka coefficient; yahi woh akela power hai jiska loop integral $\oint (z-a{)}^{k}dz$ $0$ ki jagah $2\pi i$ deta hai.

True sirf tab agar deepest term $(z-a{)}^{-3}$ hai aur $c_{-3}\ne 0$; pole order woh most-negative power hai jo present hai, sirf terms ki count nahi.

True — uniqueness sirf per annulus hold karti hai; alag singularities ke beech alag rings ek hi $f$ ke liye genuinely alag coefficient lists deti hain.

False — ek pole $r=0$ deta hai (ring ka hole point $a$ tak shrink ho jaata hai), isliye region ek punctured disk $0<|z-a|<R$ hoti hai, jo phir bhi annulus hai lekin koi hole radius nahi hai.

False — yeh poore punctured plane $0<|z|<\infty$ par converge karti hai; koi outer fence nahi hai kyunki $e^{1/z}$ ki koi doosri singularity nahi hai.

False — negative powers $a$ ke paas badhte hain lekin series is tarah design ki gayi hai ki uska sum us finite-behaving-away function ke barabar rahe; yeh sirf $a$ par diverge karti hai, jo annulus exclude karta hai.

Galat direction — $\sum z^n$ ko $|z|<1$ chahiye. $|z|>1$ mein badi quantity factor out karo: $\frac{1}{z}\cdot \frac{1}{1-1/z}={\sum }_{n\ge 0}{z}^{-n-1}$, jo $1/z$ mein ek Geometric series hai.

$0$ is $f$ ki singularity nahi hai, isliye $0$ par koi residue nahi; "$0$ ke baare mein Laurent series" ka sawaal is baare mein hai ki tum kis annulus mein expand kar rahe ho, residue ke baare mein nahi.

Ulta hai — woh non-negative powers hain (analytic part). Principal part sirf $\frac{1}{z^2}+\frac{1}{z}$ hai, yaani negative powers.

Koi "infinite-order pole" nahi hota. Infinitely many negative terms by definition ek essential singularity hoti hai — yeh kisi bhi finite-order pole se alag, zyada buri category hai."

$z=1$ par singularity ab ring ke andar hai, isliye $\frac{-1}{z-1}$ ko $1/z$ mein expand karna hoga (negative powers), jabki $\frac{1}{z-2}$ $z$ ki positive powers mein rehti hai — har piece ko apni fence ki taraf expand karo.

Annulus ke andar $a$ ke around koi bhi positively-oriented loop jo ek baar ghume woh kaam karega — deformation invariance (Cauchy integral formula) ka matlab hai ki shape irrelevant hai, sirf winding matter karti hai.

Analytic part ${\sum }_{n\ge 0}{c}_n(z-a{)}^n$ andar converge karti hai $|z-a|<R$ ke; principal part bahar converge karti hai $|z-a|>r$ ke.

Kyunki ek hi $f$ alag rings par alag expansions deta hai; annulus ka naam liye bina coefficients ambiguous hain, toh answer meaningless hai.

Kyunki ${\oint }_{\gamma }(z-a{)}^{k}dz=0$ har $k\ne -1$ ke liye aur sirf $k=-1$ ke liye $2\pi i$ milta hai, isliye loop integral sirf residue term ko filter karke baaki sab powers ko hataa deta hai — yahi Residue theorem ka engine hai.

Ek pole sirf $a$ par singularity hai aur andar kuch aur kareeb nahi; ring ka hole ek single point $a$ tak shrink ho sakta hai, jo punctured disk $0<|z-a|<R$ deta hai.

Analytic part ek genuine power series hai, isliye woh bahar ki sabse kareeb fence tak converge karti hai; bahar jaate hue jo pehli singularity milti hai woh convergence rok deti hai aur $R$ set karti hai.

Numerator $e^z$ analytic hai, isliye uski Taylor series available hai; $z^2$ se divide karne par har power do se neeche shift ho jaati hai, instantly negative-power (principal) part produce kar deta hai.

Taylor series mein sirf non-negative powers hoti hain, jo $z\to a$ par finite rehti hain; lekin pole $f$ ko blow up kara deta hai, isliye finite-behaving terms ka koi combination usse match nahi kar sakta — negative powers chahiye.

Yeh ek removable singularity hai: Laurent series mein koi negative powers nahi hain, isliye $f(a)=c_0$ redefine karne par yeh analytic ho jaata hai — "hole" cosmetic tha.

Teen — $|z|<1$, ring $1<|z|<2$, aur $|z|>2$ — fences ke beech har gap ke liye ek region.

Purely negative powers of $z$ (saari singularities ab "andar" hain), isliye poori cheez principal part hai aur large $z$ ke liye $\sim 1/z^2$ jaisi behave karti hai.

Essential singularity — infinitely many negative terms; $a$ ke paas function wildly erratic hai aur (Picard) almost har value hit karta hai.

Haan — jab koi outer singularity nahi hoti (jaise $e^{1/z}$ ya $e^z/z^2$), toh ring infinity tak extend karti hai, jo $0<|z-a|<\infty$ deta hai.

Agar do singularities coincide kar jayein toh genuinely koi gap nahi hoga, lekin distinct fences ke liye $r<R$ hamesha hota hai, isliye ek real annulus exist karti hai; ek "empty" ring batati hai ki tumne inner vs outer fence galat label kiya.