4.10.1
Ek complex function f ( z ) ( x , y ) zyaadatar maps boring hote hain (bas do real functions ko chipka diya). Special wale — jinhe analytic (ya holomorphic) kehte hain — woh hote hain jinki derivative ek single complex number ke roop mein exist karti hai, chahe aap kisi bhi direction se approach karo. Yeh ek demand bahut hi restrictive hai, aur yeh real aur imaginary parts ke beech ek hidden coupling force karti hai: Cauchy–Riemann equations .
WHAT: Ek function f : C → C f ( z ) = u ( x , y ) + i v ( x , y ) z = x + i y WHY it matters: Analytic functions plane ke "smooth, structure-preserving" maps hain. Yeh angles preserve karte hain (conformal), Laplace's equation satisfy karte hain, aur infinitely differentiable hote hain + locally power series se diye jaate hain. Yeh contour integration, residues, aur physics (fluid flow, electrostatics, signal processing) ka darwaaza hai.HOW we test analyticity: Cauchy–Riemann (CR) equations + partials ki continuity check karo.
Definition Complex differentiabilityf z 0 f ′ ( z 0 ) = lim Δ z → 0 Δ z f ( z 0 + Δ z ) − f ( z 0 ) direction se independent ho jisme Δ z → 0 f f analytic (holomorphic) hai.
Intuition Derivation kyun zaroori haiPura jaadu "direction-independence" mein hai. Chaliye force karte hain ki do alag directions mein derivative agree kare, aur dekhte hain ki equations khud nikal aayen.
f ( z ) = u ( x , y ) + i v ( x , y ) Δ z = Δ x + i Δ y
Step 1 — Real axis ke along approach karo (Δ y = 0 Δ z = Δ x Yeh step kyun? Yeh isolate karta hai ki f x f ′ ( z ) = lim Δ x → 0 Δ x u ( x + Δ x , y ) + i v ( x + Δ x , y ) − u − i v = ∂ x ∂ u + i ∂ x ∂ v .
Step 2 — Imaginary axis ke along approach karo (Δ x = 0 Δ z = i Δ y Yeh step kyun? Same derivative, lekin hum orthogonal direction mein slide karte hain. Note karo 1/ ( i Δ y ) = − i /Δ y f ′ ( z ) = lim Δ y → 0 i Δ y u ( x , y + Δ y ) + i v ( x , y + Δ y ) − u − i v = i 1 ( ∂ y ∂ u + i ∂ y ∂ v ) = ∂ y ∂ v − i ∂ y ∂ u .
Step 3 — Dono results ko equal maango.
Yeh step kyun? Yahi literally "derivative independent of direction" ka matlab hai.
real ∂ x ∂ u + i imag ∂ x ∂ v = real ∂ y ∂ v − i imag ∂ y ∂ u .
Real parts aur imaginary parts alag-alag match karo:
Intuition CR equations geometricallykya kehti hain
Yeh "ek single complex number se multiplication" ka algebraic fingerprint hain. ( x , y ) ↦ ( u , v ) J = ( u x v x u y v y ) = ( a b − b a ) , rotation + scaling hai (a + ib
CR equations akele necessary hain lekin sufficient nahin. Clean theorem yeh hai:
Definition Analyticity ke liye sufficient conditionAgar u , v continuous first partial derivatives z 0 aur wahan CR equations satisfy hoti hain, toh f z 0
Worked example Example 1 —f ( z ) = z 2 z 2 = ( x + i y ) 2 = x 2 − y 2 + 2 i x y u = x 2 − y 2 , v = 2 x y
u x = 2 x , v y = 2 x u x = v y Kyun? Pehli CR equation check karta hai.u y = − 2 y , v x = 2 y ⇒ u y = − v x Kyun? Doosri CR equation check karta hai.
Partials polynomials hain → har jagah continuous, toh f C f ′ ( z ) = u x + i v x = 2 x + i 2 y = 2 z Yeh match kyun karta hai? Complex formula d z d z 2 = 2 z
Worked example Example 2 —f ( z ) = z ˉ = x − i y kahin bhi analytic nahin hai
u = x , v = − y u x = 1 v y = − 1 Yeh fatal kyun? Har point par u x = v y z ˉ
Worked example Example 3 —f ( z ) = e z e z = e x ( cos y + i sin y ) u = e x cos y , v = e x sin y
u x = e x cos y = v y u y = − e x sin y = − v x f ′ ( z ) = u x + i v x = e x cos y + i e x sin y = e z Yeh achha kyun hai? Bilkul real exponential ki self-derivative ko mirror karta hai.
Worked example Example 4 —f ( z ) = ∣ z ∣ 2 = x 2 + y 2 u = x 2 + y 2 , v = 0 u x = 2 x = v y = 0 ⇒ x = 0 u y = 2 y = − v x = 0 ⇒ y = 0 sirf z = 0 hold karta hai. Yeh subtle kyun hai? f 0 kahin bhi analytic nahin , kyunki analyticity ko ek open set par differentiability chahiye, aur ek single point open nahin hota.
Common mistake "CR ek point par hold karta hai ⇒ wahan analytic hai."Yeh sahi kyun lagta hai: Humne CR seedha derivative limit se derive kiya, toh yeh equivalent lagta hai. Flaw: CR sirf necessary hai; aapko continuous partials bhi chahiye (ya ek open neighborhood par differentiability). Example 4, CR sirf 0 Fix: CR ko ek region par check karo aur confirm karo ki partials continuous hain.
Common mistake Minus sign bhulaana:u y = v x Yeh sahi kyun lagta hai: Symmetry — aap expect karte hain ki dono equations same dikhein. Flaw: Δ z = i Δ y i 1/ i = − i Fix: yaad rakho "cross terms get a minus": u x = + v y u y = − v x
Common mistake Yeh sochna ki plane ka koi bhi smooth real map analytic hai.Yeh sahi kyun lagta hai: "x y Flaw: Real differentiability ko sirf real Jacobian exist karna chahiye; complex differentiability demand karti hai ki Jacobian ek rotation–scaling matrix ( a b − b a ) Fix: analytic ⇒ Jacobian special form ⇒ CR.
"Right pairs match, cross pairs clash."
Right (same variable order) u x = v y match karta hai; cross u y = − v x clash karta hai (minus sign).
Ya: U p-X equals V -Y , aur doosra diagonal negative carry karta hai.
Recall Feynman: ek 12-saal ke bachche ko samjhaoSocho ek stretchy rubber sheet jisme arrows bane hain (ek map). Ek "achha" map sirf chhote neighborhoods ko spin aur zoom kar sakta hai — woh unhe twist aur bada kar sakta hai, lekin saare angles same rakhne chahiye aur kabhi andar-bahar flip nahin karna chahiye (koi mirror nahin). Cauchy–Riemann equations bas woh secret handshake hai jo kehta hai "haan, yeh map sirf spins-aur-zooms karta hai." Agar koi map cheezein flip karta hai (jaise ek mirror), ya upar-neeche se zyaada sideways stretch karta hai, toh yeh handshake tod deta hai aur "achha" nahin hai.
f z 0 Limit Δ z f ( z 0 + Δ z ) − f ( z 0 ) Δ z → 0
Cauchy–Riemann equations batao. u x = v y u y = − v x f = u + i v
CR equations kyun aati hain? Real axis ke along compute ki gayi f ′ u x + i v x v y − i u y
Kya CR equations analyticity ke liye sufficient hain? Nahin — sirf necessary hain. Aapko u , v
Ek aisa function batao jo exactly ek point par CR satisfy kare lekin kahin bhi analytic na ho. f ( z ) = ∣ z ∣ 2 = x 2 + y 2 z = 0
f ( z ) = z ˉ u = x , v = − y u x = 1 = v y = − 1
Analytic function ke real aur imaginary parts kaunsi PDE satisfy karte hain? Laplace's equation u xx + u y y = 0 v
Partials ke terms mein f ′ ( z ) f ′ = u x + i v x = v y − i u y
Analyticity se forced Jacobian form kya hai? ( a b − b a ) a + ib
f ( z ) = e z f ′ e z u = e x cos y , v = e x sin y f ′ = u x + i v x = e x cos y + i e x sin y = e z
derivative independent of direction
approach along imaginary axis
Complex differentiability
Contour integration & residues
Fluid flow & electrostatics
