Complex analysis — analytic functions, Cauchy-Riemann equations: Visual walkthrough

4.10.1 · D2 · Maths › Advanced Topics (Elite Level) › Complex analysis — analytic functions, Cauchy-Riemann equati

Step 0 — Complex function ki picture (setup, abhi koi formula nahi)

Figure dekho: left plane mein input point $z$ hai; ek arrow jis par $f$ likha hai usse right plane mein le jaata hai, jahan woh output ke roop mein coordinates $(u, v)$ ke saath land karta hai. Hum jo bhi karte hain woh is baare mein hai ki jab hum input point ko thoda sa hilate hain toh woh landing spot kaise shift hota hai.

Step 1 — "Derivative" ka matlab yahan kya hai

Figure mein input point ko kai colored arrows $Δ z$ alag-alag directions mein point karte hue dikhaya gaya hai. Complex differentiability maangti hai ki un sab ko, $f$ se push karne aur divide karne ke baad, wahi identical $f^{'} (z)$ milna chahiye.

Step 2 — Real axis ke saath chalte hain (horizontal approach)

$Δ z = Δ x$ set karo (ek real number, purely horizontal). Tab: $f^{'} (z) = lim_{Δ x \to 0} \frac{u ( x + Δ x , y ) + i v ( x + Δ x , y ) - u - i v}{Δ x} .$

Fraction ko uske right-part aur up-part mein split karo — har ek ordinary rate of change ban jaata hai: $f^{'} (z) = u_{x} Δ x \to 0 lim \frac{u ( x + Δ x , y ) - u}{Δ x} + i v_{x} Δ x \to 0 lim \frac{v ( x + Δ x , y ) - v}{Δ x} .$

Figure mein nudge (mint arrow) purely right point karta hai. Do side-graphs $u$ aur $v$ ko un raston ke roop mein dikhate hain jinhe tum $x$ mein chalte ho; unki slopes $u_{x}$ aur $v_{x}$ hain — derivative ke real aur imaginary halves is tarah measured.

Step 3 — Imaginary axis ke saath chalte hain (vertical approach)

Ab $Δ z = i Δ y$ set karo (purely vertical). Key algebra fact: $\frac{1}{i} = - i kyunki i \cdot (- i) = - i^{2} = - (- 1) = 1.$

Compute karo: $f^{'} (z) = lim_{Δ y \to 0} \frac{u ( x , y + Δ y ) + i v ( x , y + Δ y ) - u - i v}{i Δ y} = \frac{1}{i} (u_{y} + i v_{y}) .$

Yahan $u_{y}, v_{y}$ slopes hain jab tum sirf upward chalte ho ( $x$ fixed rakhke). $\frac{1}{i} = - i$ use karke multiply karo: $\frac{1}{i} (u_{y} + i v_{y}) = - i u_{y} - i^{2} v_{y} = real v_{y} - i imag u_{y} .$

Figure Step 2 ka setup repeat karta hai lekin ek coral arrow seedha upar point karta hua, aur highlight karta hai ki $1/ i$ turn chhota output arrow ko ek quarter-turn clockwise kaise rotate karta hai — woh rotation exactly wahi hai jo $u_{y}, v_{y}$ ko $v_{y} - i u_{y}$ mein rearrange karta hai.

Step 4 — Dono answers ko agree hone par majboor karo

$real, A se u_{x} + i imag, A se v_{x} = real, B se v_{y} - i imag, B se u_{y} .$

Real parts match karo, phir imaginary parts: $real : u_{x} = v_{y}, imag : v_{x} = - u_{y} .$

Figure do derivative arrows (mint = horizontal result, coral = vertical result) ko overlay karta hai aur dikhata hai ki CR hold hone par woh ek arrow mein collapse ho jaate hain. Do component-matchings exactly wahan label hain jahan arrows align karte hain.

Step 5 — Equations ka geometric matlab kya hai (special Jacobian)

Chaar partials stack karo: $J = (u_{x} v_{x} u_{y} v_{y}) u_{x} = v_{y} u_{y} = - v_{x} (a b - b a), a := u_{x}, b := v_{x} .$

Figure ek tiny square grid ko is matrix se push hote dikhata hai: woh rotated aur uniformly resized hokar niklata hai — corners abhi bhi square hain, orientation preserved hai. Yeh "angles survive" property hi reason hai ki analytic maps conformal hote hain (dekho Conformal mappings).

Step 6 — Degenerate case: ek akele point par CR kaafi nahi hai

$f (z) = ∣ z ∣^{2} = x^{2} + y^{2}$ lo, toh $u = x^{2} + y^{2}$ , $v = 0$ . $u_{x} = 2 x, v_{y} = 0 \Rightarrow u_{x} = v_{y} ke liye x = 0 chahiye; u_{y} = 2 y, v_{x} = 0 \Rightarrow u_{y} = - v_{x} ke liye y = 0 chahiye .$

Dono sirf single point $z = 0$ par hold karte hain.

Figure $z = 0$ ko mark karta hai akele jagah ke roop mein jahan do component conditions ( $x = 0$ line aur $y = 0$ line) intersect karti hain — ek lone point, open patch nahi. Analyticity ko ek full shaded disc chahiye; yahan hume sirf ek single pixel milta hai.

Step 7 — Degenerate case: ek reflection har jagah handshake tod deta hai

$f (z) = \overset{z}{ˉ} = x - i y$ lo, toh $u = x$ , $v = - y$ . $u_{x} = 1, v_{y} = - 1 \Rightarrow u_{x} \neq = v_{y} (har point par fail).$

Figure ek grid ko $\overset{z}{ˉ}$ se push karta hai: woh mirror-flipped hokar niklata hai (ek letter "R" apni mirror image ban jaata hai). Koi bhi rotate-and-scale yeh nahi kar sakta, toh special Jacobian impossible hai, aur CR har point par toota hua hai — kahin bhi analytic nahi.

Ek-picture summary

Summary ko left se right padhein: horizontal se nudge karo ( $\to u_{x} + i v_{x}$ ) aur vertical se ( $\to v_{y} - i u_{y}$ ); $1/ i$ quarter-turn (upar) wahi hai jo doosre ko rearrange karta hai. Unhe equal karne par do boxed CR equations nikalti hain, jo Jacobian ko rotate-and-scale form mein reshape karti hain — ek map jo spin aur zoom karta hai lekin kabhi flip nahi karta.

Recall Feynman retelling — poora walkthrough simple shabdon mein

Ek stretchy sheet ki picture karo jisme ek point hai. Main us point ko thoda sa right mein poke karta hun aur dekhta hun ki map use kahan bhejta hai — yeh mujhe ek "rate" deta hai, do numbers se bana ( $u_{x}$ aur $v_{x}$ ). Phir main usi point ko thoda sa upward poke karta hun aur phir dekhta hun — yeh doosra rate deta hai. Lekin upward poke karna ek quarter-turn chura leta hai (woh $1/ i$ hai), jo do numbers ko idhar-udhar shuffle karta hai aur ek sign flip karta hai, toh yeh rate $v_{y}$ aur $- u_{y}$ ki tarah padhta hai.

Ab yahan game ka rule hai: ek "nice" map ko chahe main kisi bhi taraf poke karun usi tarah zoom-and-spin karna hoga. Toh rightward answer aur upward answer identical complex number hone chahiye. Right-coordinates equal karne se $u_{x} = v_{y}$ milta hai; up-coordinates equal karne se $u_{y} = - v_{x}$ milta hai. Woh do lines hain Cauchy–Riemann equations, aur minus sign bas us quarter-turn ka bacha hua fingerprint hai.

Pictures se do warnings: sirf right-aur-up match karna hamesha ek akele dot par kaafi nahi hota (jaise $∣ z ∣^{2}$ origin par) — tumhe puri patch par chahiye. Aur ek mirror map (jaise $\overset{z}{ˉ}$ ) sheet ko ulta flip kar deta hai, jo koi bhi spin-and-zoom nahi kar sakta, toh woh har jagah flunk karta hai.

Recall

Real axis ke saath $f^{'} (z)$ kya equal hota hai? ::: $u_{x} + i v_{x}$ Imaginary axis ke saath $f^{'} (z)$ kya equal hota hai? ::: $v_{y} - i u_{y}$ $u_{y} = - v_{x}$ mein minus sign kahan se aata hai? ::: Vertical approach mein $1/ i = - i$ se — quarter-turn se. $f (z) = ∣ z ∣^{2}$ analytic kyun nahi hai even though CR $0$ par hold karta hai? ::: Ek single point open region nahi hoti; analyticity ko puri neighbourhood par CR chahiye. Jab CR hold karta hai toh Jacobian kya shape leta hai? ::: $(a b - b a)$ — ek rotation-and-scaling ( $a + ib$ se multiply).

Active recall

Do directions kaun se hain jinhe hum CR derive karne ke liye compare karte hain, aur har ek kya derivative deta hai?

Horizontal (

Δ z = Δ x

) deta hai

u_{x} + i v_{x}

; vertical (

Δ z = i Δ y

) deta hai

v_{y} - i u_{y}

Cauchy–Riemann equations state karo.

u_{x} = v_{y}

aur

u_{y} = - v_{x}

Ek complex equation do real mein kyun split ho sakti hai?

Do complex numbers equal hote hain tabhi jab dono real aur imaginary parts alag-alag match karein.

CR-Jacobian kaunsi geometric transformation represent karta hai?

Ek pure rotation-and-scaling (ek single complex number

a + ib

se multiplication) — orientation-preserving.