Complex analysis — analytic functions, Cauchy-Riemann equations: Question bank

4.10.1 · D5 · Maths › Advanced Topics (Elite Level) › Complex analysis — analytic functions, Cauchy-Riemann equati

Definition Notation aur words jis par ye page rely karta hai (isse pehle kuch nahi use hua)

$z = x + i y$ plane ka ek point hai; $f (z) = u (x, y) + i v (x, y)$ output ko ek real part $u$ aur ek imaginary part $v$ mein split karta hai.
$u_{x}$ jaisa subscript matlab hai " $u$ ka $x$ ke respect mein partial derivative" — kitni jaldi $u$ change hota hai jab tum sirf $x$ ko thoda sa nudge karo, $y$ ko fixed rakh ke.
$\overset{z}{ˉ} = x - i y$ real axis ke paar $z$ ki mirror image hai.
$Δ z$ $z$ se ek choti si displacement hai: tum $z$ se $z + Δ z$ tak step karte ho. Kyunki $Δ z$ khud ek complex number hai, $Δ z = Δ x + i Δ y$ ek horizontal step $Δ x$ aur ek vertical step $Δ y$ carry karta hai. "Real axis ke saath" approach karna matlab $Δ y = 0$ ; "imaginary axis ke saath" matlab $Δ x = 0$ toh $Δ z = i Δ y$ .
Complex derivative $f^{'} (z)$ ek single complex number hai $f^{'} (z) = lim_{Δ z \to 0} \frac{f ( z + Δ z ) - f ( z )}{Δ z},$ jise tiny step $Δ z$ ki direction se koi bhi fark nahi padna chahiye — same value dena chahiye. Jab ye exist karta hai, do axis directions par kaam karne se (parent note dekho) ye twin formulas milte hain: $f^{'} (z) = u_{x} + i v_{x} = v_{y} - i u_{y}$ . $f^{'} (z) = a + ib$ likhne par, numbers $a = u_{x} = v_{y}$ aur $b = v_{x} = - u_{y}$ us ek derivative ke real aur imaginary parts hain — ye wahi $a, b$ hain jo neeche har figure aur matrix mein hain.
Open set: ek region jisme koi boundary points nahi hote, toh har point ke around ek poori tiny disk hoti hai jo set ke andar hoti hai. Ek point ka neighborhood uske around ek aisi choti disk hai. Kisi set ka interior wo saare points hain jinke paas set ke andar ek neighborhood hai. Ek single point ${0}$ ka empty interior hai — ek dot ke andar koi disk fit nahi ho sakti. Ek region par Analytic matlab har point par ek open set mein differentiable hona; ek point par differentiable hona isse weaker hai.

Figure — Complex analysis — analytic functions, Cauchy-Riemann equations

Upar wali picture wahi poori wajah hai jis ke liye CR exist karta hai: ek analytic map, close up dekha jaye, toh ek choti disk ko sirf spin aur zoom kar sakta hai (left/red); wo use kabhi flip nahi kar sakta jaise mirror karta hai (right). Is page ka baaki hissa har us tarike ko dhundhta hai jisme students un do words mein se kisi ek ko bhool jaate hain.

True or false — justify

(x, y) \mapsto (u, v)

map ka real Jacobian hona (dono partials exist karna) matlab hai

f

complex-differentiable hai.

False — real differentiability sirf ye maangti hai ki ek linear approximation exist kare; complex differentiability additionally force karti hai ki wo linear map ek rotation-plus-scaling ho, yaani Jacobian ka ek special form hona zaroori hai

(a b - b a)

jahan

a = u_{x} = v_{y}

aur

b = v_{x} = - u_{y}

(

f^{'} (z) = a + ib

ke real aur imaginary parts), jo exactly CR hai.

Agar

f

aur

g

ek region par analytic hain, toh

f \cdot g

wahan analytic hai.

True — complex derivatives ke liye product rule bilkul real case ki tarah kaam karta hai, toh complex limit

(f g)^{'} = f^{'} g + f g^{'}

region ke har point par exist karta hai; analyticity products ke under closed hai.

Agar

f

ek region par analytic aur nonzero hai, toh

1/ f

wahan analytic hai.

True — quotient complex-differentiable hai jahan bhi denominator nonzero hai, aur humne assume kiya tha

f \neq = 0

, toh koi singularity appear nahi hoti;

(1/ f)^{'} = - f^{'} / f^{2}

Poore

C

par analytic function (entire) necessarily ek polynomial hona chahiye.

False —

e^{z}

sin z

cos z

entire hain lekin polynomials nahi hain; "entire" ka sirf matlab hai ki complex derivative har jagah exist karta hai, jo infinite radius of convergence wali power series bhi satisfy karti hain (dekho Power series and radius of convergence).

Agar

u

aur

v

dono harmonic hain, toh

f = u + i v

automatically analytic hai.

False — dono ka harmonic hona necessary hai lekin sufficient nahi; unhe harmonic conjugates hona chahiye (CR se linked:

u_{x} = v_{y}

u_{y} = - v_{x}

). Do alag harmonic functions ko ek saath jodne par generally CR break ho jaata hai.

Real part

u

akela (ek constant tak) simply-connected region par apna harmonic conjugate

v

determine karta hai.

True — CR deta hai

v_{x} = - u_{y}

aur

v_{y} = u_{x}

, toh

v

ko in known partials ko integrate karke recover kiya jaata hai; ek additive real constant of integration ki hi freedom hoti hai.

Agar ek analytic map ke kisi point par

f^{'} (z_{0}) = 0

hai, toh map wahan bhi angles preserve karta hai.

False — conformality ke liye complex derivative

f^{'} (z_{0}) \neq = 0

chahiye; jahan ye vanish karta hai wahan local rotation-scaling collapse ho jaata hai aur angles multiply ho sakte hain (jaise

z^{2}

zero

0

par angles double karta hai). Dekho Conformal mappings.

Ek analytic function jiska real part

u

ek region par constant hai woh khud wahan constant hona chahiye.

True —

u

constant hone se

u_{x} = u_{y} = 0

milta hai; CR phir

v_{y} = 0

aur

v_{x} = 0

force karta hai, toh

v

bhi constant hai, isliye

f

constant hai.

Agar

∣ f (z) ∣

ek region par constant hai aur

f

analytic hai, toh

f

constant hai.

True — constant modulus matlab

u^{2} + v^{2}

constant hai; differentiate karke aur CR apply karne par saare partials vanish ho jaate hain (unless

∣ f ∣ = 0

ho, jisse

f \equiv 0

milta hai), toh

f

constant hai.

Har function jo CR equations ek single point par satisfy karta hai, woh us point par complex-differentiable hai.

False — CR ek point par necessary hai lekin partials ki continuity ke bina sufficient nahi; aisa function construct kiya ja sakta hai jo

0

par CR satisfy kare phir bhi wahan differentiable na ho kyunki diagonal approaches ke saath directional limits disagree karte hain.

Spot the error

f (z) = \overset{z}{ˉ}

ke partials hain

u_{x} = 1

v_{y} = - 1

, lekin kyunki

u_{y} = 0 = - v_{x}

toh doosra CR equation hold karta hai, toh

f

'half analytic' hai."

Koi "half analytic" nahi hota — analyticity ke liye dono CR equations simultaneously chahiye; pehla fail ho raha hai (

u_{x} = 1 \neq = - 1 = v_{y}

), toh

f = \overset{z}{ˉ}

kahin bhi analytic nahi hai, bas itna.

"Mainne CR ko

u_{x} = v_{y}

aur

u_{y} = v_{x}

likha; Jacobian ke dono diagonals match karte hain, clean aur symmetric."

Minus sign missing hai: sahi pair hai

u_{y} = - v_{x}

. Yaad karo

Δ z = Δ x + i Δ y

; imaginary axis ke saath approach karne par

Δ z = i Δ y

milta hai, aur isse divide karne par

1/ i = - i

use hota hai — wahi factor exactly woh jagah hai jahan sign paida hota hai. Symmetric dikhne wale equations ek reflection describe karte hain, rotation nahi.

f (z) = ∣ z ∣^{2}

z = 0

par CR satisfy karta hai, isliye ye

0

par analytic hai."

Analyticity ke liye ek open set par differentiability chahiye, aur

{0}

ek single point hai jiska empty interior hai — ek bhi disk ek dot ke andar fit nahi hoti.

f

sirf

0

par complex-differentiable hai; koi aise neighborhood nahi hai jahan CR hold kare, toh ye kahin bhi analytic nahi hai.

"Kyunki

u = x^{2} + y^{2}

ek nice smooth polynomial hai,

f = ∣ z ∣^{2}

ek nice analytic function hai."

Real functions ke roop mein

u, v

ki smoothness apne aap mein irrelevant hai; tumhe unhe CR ke through feed karna hoga. Yahan

v = 0

force karta hai

u_{x} = u_{y} = 0

, jo sirf origin par true hai, toh smoothness analyticity ko kabhi rescue nahi kar sakti.

"CR do equations hain, toh

u

jaanne se

v

uniquely pin down ho jaata hai, koi freedom nahi."

v

ke partials pin down karta hai, isliye

v

ek additive real constant tak — integration hamesha ek constant chodti hai. Aur ek non-simply-connected region par existence bhi fail ho sakti hai.

f^{'} (z) = u_{x} + i v_{x}

aur saath hi

f^{'} (z) = v_{y} - i u_{y}

; ye do alag derivatives hain, toh

f

ke do derivatives hain."

Ye ek hi single complex derivative ke liye do formulas hain — inhe equate karna exactly hai jaise CR arise hota hai. Jab CR hold karta hai, dono expressions numerically identical hote hain.

Why questions

"Direction-independence" demand karna ek limit ka do real equations kyun produce karta hai?

Ek complex equation ek real aur ek imaginary component pack karta hai; real-axis derivative

u_{x} + i v_{x}

ko imaginary-axis derivative

v_{y} - i u_{y}

ke equal set karne par real parts match karne force hoti hain (

u_{x} = v_{y}

) aur imaginary parts match karne force hoti hain (

u_{y} = - v_{x}

) — ek complex demand se do constraints.

Ek analytic map ka Jacobian

(a b - b a)

jaisa kyun dikhna chahiye?

Kyunki local action hai "single complex number

f^{'} (z) = a + ib

se multiply karo" (jahan

a = u_{x} = v_{y}

b = v_{x}

), aur

a + ib

se multiply karna,

2 \times 2

real matrix form mein, exactly wahi rotation-scaling matrix hai — CR sirf ye matrix condition hai jo likhit roop mein likhi hui hai. Dekho Jacobian and the multivariable chain rule.

Analyticity real aur imaginary parts ko Laplace's equation satisfy karne par kyun force karti hai?

Analytic functions infinitely differentiable nikle hain, toh

u, v

C^{2}

hain (unke second partials exist hain aur continuous hain); wahi continuity exactly woh hai jo mixed partials ki equality

v_{y x} = v_{x y}

(Clairaut's theorem) ko license deti hai. CR ko cross-differentiate karne par milta hai

u_{xx} = v_{y x}

aur

u_{y y} = - v_{x y}

, aur inhe add karne par

v

terms cancel ho ke

u_{xx} + u_{y y} = 0

bachta hai. Dekho Harmonic functions and Laplace's equation.

Analytic maps ke liye ek reflection (jaise

\overset{z}{ˉ}

) forbidden kyun hai?

Reflections orientation reverse karti hain — unke Jacobian ka determinant

- (a^{2} + b^{2}) < 0

hota hai, jo kabhi bhi ek rotation-scaling ke

+ (a^{2} + b^{2})

ke equal nahi ho sakta; CR orientation-preservation encode karta hai, toh mirrors ise har jagah break karte hain.

Sufficiency ke liye partials ke sirf existence par nahi, continuity par bhi insist kyun karte hain?

Partials ka existence sirf do axis directions control karta hai; continuity hi guarantee karti hai ki linear approximation har direction ke saath valid hai, "CR hold karta hai" aur "

\frac{f ( z + Δ z ) - f ( z )}{Δ z}

ka poora directional limit exist karta hai aur agree karta hai" ke beech ka gap close karti hai.

Ek analytic function ka derivative

Δ z

kis direction se approach kare is par depend kyun nahi karta, jabki ek generic real map ka karta hai?

Ek generic real map ke liye

Δ z

ki alag alag directions alag rates deti hain (directional derivative vary karta hai); sirf jab Jacobian ek pure rotation-scaling hota hai tabhi har direction ko same factor se scale kiya jaata hai aur same angle se rotate kiya jaata hai — wahi single common factor hai

f^{'} (z)

Angles preserve karna ("conformal") ek analytic function ka nonzero derivative ke saath hona kyun equivalent hai?

f^{'} (z) = r e^{i θ}

se multiply karna har local vector ko same

θ

se rotate karta hai aur same

r

se scale karta hai; saari directions ka equal rotation kisi bhi do curves ke beech ke angle ko unchanged rakhta hai. Agar

f^{'} = 0

toh local picture degenerate ho jaata hai aur ye fail ho jaata hai.

Edge cases

Kya ek function jo exactly ek isolated point par complex-differentiable hai, kahan bhi analytic hai?

Nahi — analyticity ke liye ek open set mein differentiability chahiye; ek akela point (ya poora curve bhi) empty interior rakhta hai, toh koi region nahi hai jis par

f

qualify kare.

Analyticity ka kya hota hai us point par jahan

f^{'} (z_{0}) = 0

ho, jaise

z^{2}

origin par?

Function wahan bhi analytic hai (ek neighborhood mein complex-differentiable), lekin us point par conformal nahi hai — angles multiply ho jaate hain (

z^{2}

ke liye, double ho jaate hain), kyunki rotation-scaling degenerate ho jaata hai jab scale factor

0

hota hai.

Kya ek nonconstant analytic function ek open region par real-valued (

v \equiv 0

) ho sakti hai?

Nahi —

v \equiv 0

deta hai

v_{x} = v_{y} = 0

, toh CR force karta hai

u_{x} = u_{y} = 0

, jisse

u

constant ho jaata hai; ek region par real-valued analytic function necessarily constant hoti hai.

Kya CR mein modification chahiye agar

f

sirf ek curve ya boundary par define hai, open set par nahi?

Haan spirit mein — complex differentiability genuinely ek 2D limit hai (

Δ z

step ko har taraf point karne ki freedom chahiye), toh tum use bina har direction se approach karne ki jagah ke state bhi nahi kar sakte; ek mere curve par analyticity ka koi notion nahi, sirf ek restricted function ka.

Kya

f (z) = \overset{z}{ˉ}^{2}

kahan bhi analytic hai?

Nahi —

\overset{z}{ˉ}^{2} = x^{2} - y^{2} - 2 i x y

deta hai

u = x^{2} - y^{2}

v = - 2 x y

, toh

u_{x} = 2 x

v_{y} = - 2 x

; CR fail ho jaata hai siwaay jahan

x = 0

ho, yaani sirf ek line par (koi open set nahi), toh kahin bhi analytic nahi.

Agar

u, v

ke partials exist karte hain aur har jagah CR satisfy karte hain lekin ek point par discontinuous hain, toh kya galat ho sakta hai?

Complex differentiability us discontinuity par fail ho sakti hai chahe CR hold kare — classic counterexample mein directional limits diagonals ke saath disagree karte hain, toh sufficiency theorem mein partials ki continuity genuinely kaam kar rahi hai.

f (z) = z \overset{z}{ˉ} = ∣ z ∣^{2}

versus

f (z) = z^{2}

ki analytic status kya hai, aur sharp contrast kyun?

z^{2}

entire hai (sirf

z

mein holomorphic), jabki

z \overset{z}{ˉ}

\overset{z}{ˉ}

par depend karta hai, aur koi bhi explicit

\overset{z}{ˉ}

-dependence measure-zero set se bahar CR violate karta hai — analytic functions exactly whi hote hain jo "

\overset{z}{ˉ}

ko nahi dekhte."

Kya do alag analytic functions ek connected region par same real part

u

share kar sakte hain?

Sirf agar woh sirf ek purely imaginary constant

i c

se differ karen — CR

v

ke partials ko

u

se fix karta hai,

v

mein sirf ek additive real constant chodta hai; toh functions

f

aur

f + i c

hain.

Recall Upar wale har trap ke liye ek-line filter

Pucho: "Kya ye claim ek open region par dono CR equations aur continuous partials ke saath survive karta hai?" Agar un teen words mein se koi bhi drop ho jaaye, toh trap suspect karo.