Complex analysis — analytic functions, Cauchy-Riemann equations: Worked examples

4.10.1 · D3 · Maths › Advanced Topics (Elite Level) › Complex analysis — analytic functions, Cauchy-Riemann equati

Hum sirf wohi notation use karte hain jo parent ne already build ki hai: $z = x + i y$ , $f (z) = u (x, y) + i v (x, y)$ , partial derivatives $u_{x} = \partial u / \partial x$ ( $u$ ka rate of change jab $x$ move kare, $y$ frozen ho), aur do CR equations. Agar yeh sab shaky lag raha hai, toh pehle parent note dobara padho.

Scenario matrix

Is topic ke har problem ka answer in cells mein se exactly ek mein aata hai. Har row ek class of behaviour hai; last column us example ka naam deta hai jo usse best represent karta hai.

Cell	Kya cheez use distinct banati hai	Example
A. Analytic everywhere (entire)	CR har point par hold karta hai; partials continuous hain	Ex 1: $z^{3}$
B. Nowhere analytic	CR har point par fail karta hai (orientation-reversing)	Ex 2: $f = \overset{z}{ˉ}^{2}$
C. Analytic except a pole	CR ek bure point ko chhod kar har jagah hold karta hai jahan $f$ blow up karta hai	Ex 3: $1/ z$
D. CR sirf ek curve par (degenerate)	CR ek line/point par hold karta hai, koi open set nahi ⇒ analytic nahi	Ex 4: $z \overset{z}{ˉ} + z$
E. Conjugate build karo	Harmonic $u$ diya gaya hai, $v$ construct karo taaki $f$ analytic ho	Ex 5: $u = x^{3} - 3 x y^{2}$
F. Polar / non-Cartesian input	Function $r, θ$ mein diya gaya hai — polar CR ki zaroorat hai	Ex 6: $lo g z$
G. Real-world word problem	Physics analyticity ke roop mein disguise hua (flow / potential)	Ex 7: fluid potential
H. Exam-style twist	"Kin $a, b$ ke liye…" — parameter ke liye solve karo	Ex 8: $f = x^{2} + a x y + b y^{2} + i (\dots)$

Signs aur quadrants ("every quadrant" ki demand) Ex 6 ke andar dikhte hain, jahan $θ$ charon quadrants mein range karta hai aur $lo g$ ki branch carefully choose karni padti hai — yahi jagah hai jahan is topic ke liye sign-of- $x$ , sign-of- $y$ logic rehti hai.

Ex 1 — Cell A: analytic everywhere

Worked example

$f (z) = z^{3}$ . Kya yeh analytic hai? Kahan? Forecast: Padhne se pehle guess karo — $z$ mein polynomial hai, toh… har jagah? Kahin nahi? Ek point par? Koi jawab commit karo.

Step 1. Real aur imaginary parts mein expand karo. $z^{3} = (x + i y)^{3} = x^{3} + 3 x^{2} (i y) + 3 x (i y)^{2} + (i y)^{3} = x^{3} - 3 x y^{2} + i (3 x^{2} y - y^{3})$ . Toh $u = x^{3} - 3 x y^{2}$ , $v = 3 x^{2} y - y^{3}$ . Yeh step kyun? CR $u$ aur $v$ ke terms mein state ki gayi hai, isliye kuch bhi test karne se pehle hume unhe physically alag karna hoga.

Step 2. Charon partials compute karo. $u_{x} = 3 x^{2} - 3 y^{2}, u_{y} = - 6 x y, v_{x} = 6 x y, v_{y} = 3 x^{2} - 3 y^{2}$ . Yeh step kyun? CR $u_{x}$ ko $v_{y}$ se aur $u_{y}$ ko $- v_{x}$ se compare karta hai; hume table par har ingredient chahiye.

Step 3. Dono CR equations check karo. $u_{x} = 3 x^{2} - 3 y^{2} = v_{y}$ ✅. $u_{y} = - 6 x y$ aur $- v_{x} = - 6 x y$ ✅. Yeh step kyun? Dono hold hone chahiye; akela ek kaafi nahi hai.

Step 4. Continuity. Charon partials polynomials hain ⇒ pure $C$ par continuous hain. Sufficiency theorem se, $f$ entire hai. Yeh step kyun? CR + ek open set par continuous partials = analytic. Continuity skip karna wahi classic error hai jiske baare mein parent ne warn kiya tha.

Verify: Complex derivative $3 z^{2}$ hona chahiye. Compute karo $f^{'} = u_{x} + i v_{x} = (3 x^{2} - 3 y^{2}) + i 6 x y = 3 (x^{2} - y^{2} + 2 i x y) = 3 (x + i y)^{2} = 3 z^{2}$ ✅. Yeh check kyun? Ek independent formula ( $\frac{d}{d z} z^{3} = 3 z^{2}$ power rule se) us value se agree karna chahiye jo hume partial derivatives ko join karke mili — agreement ka matlab hai koi algebra slip nahi hui. Units-style sanity: ek length-2 vector ko cube karna uski "spin rate" ko triple karta hai — $3$ ke saath consistent hai.

Ex 2 — Cell B: nowhere analytic

Worked example

$f (z) = \overset{z}{ˉ}^{2} = (x - i y)^{2}$ . Forecast: Yeh ek square hai, aur $z^{2}$ entire tha. Kya bar isse har jagah barbaad kar deta hai, ya sirf ek curve par?

Step 1. Expand karo. $\overset{z}{ˉ}^{2} = (x - i y)^{2} = x^{2} - y^{2} - 2 i x y$ . Toh $u = x^{2} - y^{2}$ , $v = - 2 x y$ . Yeh step kyun? Wahi reason jaise hamesha — CR ko $u, v$ alag chahiye.

Step 2. Partials: $u_{x} = 2 x, u_{y} = - 2 y, v_{x} = - 2 y, v_{y} = - 2 x$ . Yeh step kyun? Do CR equations $u_{x}$ ko $v_{y}$ se aur $u_{y}$ ko $- v_{x}$ se compare karti hain, isliye koi bhi equation test karne se pehle hume charon partials haath mein chahiye.

Step 3. Pehli CR: $u_{x} = 2 x$ vs $v_{y} = - 2 x$ . Equal tabhi jab $2 x = - 2 x \Rightarrow x = 0$ . Doosri CR: $u_{y} = - 2 y$ vs $- v_{x} = 2 y$ . Equal tabhi jab $y = 0$ . Yeh step kyun? Hum wo set dhundh rahe hain jahan DONO hold karein.

Step 4. Dono sirf $x = 0$ aur $y = 0$ par hold karte hain, yaani single point $z = 0$ . Ek single point ek open set nahi hai ⇒ nowhere analytic. Yeh step kyun? Analyticity ke liye ek open neighborhood chahiye; ek point kabhi yeh provide nahi kar sakta.

Verify: Geometrically $\overset{z}{ˉ}^{2}$ pehle reflect karta hai (bar se) phir square karta hai. Reflection orientation reverse karta hai, "spin-and-zoom only" handshake tod deta hai — "nowhere analytic" se match karta hai. Yeh check kyun? Ek geometric argument jo algebra se agree kare, partials mein sign slip ke against guard karta hai. $z = 1$ par numeric spot-check ( $x = 1, y = 0$ ): $u_{x} = 2, v_{y} = - 2$ , unequal ✅ (CR genuinely origin se door fail karta hai).

Ex 3 — Cell C: analytic except one pole

Worked example

$f (z) = \frac{1}{z}$ . Yeh kahan analytic hai? Forecast: $z$ se division — $z = 0$ par trouble, lekin baaki har jagah kya?

Step 1. Rationalize karo: $\frac{1}{z} = \frac{z ˉ}{z z ˉ} = \frac{x - i y}{x ^{2} + y ^{2}}$ . Toh $u = \frac{x}{x ^{2} + y ^{2}}$ , $v = \frac{- y}{x ^{2} + y ^{2}}$ , valid jahan bhi $x^{2} + y^{2} \neq = 0$ , yaani $z \neq = 0$ . Yeh step kyun? Hum CR tabhi test kar sakte hain jahan $u, v$ actually defined hों; domain shuru se $0$ ko exclude karta hai.

Step 2. Maan lo $D = x^{2} + y^{2}$ . Quotient rule use karke: $u_{x} = \frac{D - x \cdot 2 x}{D ^{2}} = \frac{y ^{2} - x ^{2}}{D ^{2}}, v_{y} = \frac{- D - ( - y ) \cdot 2 y}{D ^{2}} = \frac{y ^{2} - x ^{2}}{D ^{2}}$ . Toh $u_{x} = v_{y}$ ✅. Yeh step kyun? Punctured plane par pehli CR equation ka direct check.

Step 3. $u_{y} = \frac{- x \cdot 2 y}{D ^{2}} = \frac{- 2 x y}{D ^{2}}$ , aur $v_{x} = \frac{- ( - y ) \cdot 2 x}{D ^{2}} = \frac{2 x y}{D ^{2}}$ , toh $- v_{x} = \frac{- 2 x y}{D ^{2}} = u_{y}$ ✅. Yeh step kyun? Doosri CR equation bhi check karni padti hai; sirf pehli pass karna exactly wahi half-check hai jiske baare mein parent ke mistake callout ne warn kiya tha.

Step 4. Partials rational functions hain, $z = 0$ ko chhod kar har jagah continuous hain. Isliye $f$ , $C ∖ {0}$ par analytic hai; point $0$ ek pole hai, domain ka part nahi. Yeh step kyun? Continuity sirf wahan fail hoti hai jahan denominator vanish karta hai; woh akela point exclude hai, isliye open-set requirement baaki har jagah meet hoti hai.

Verify: $f^{'}$ , $- 1/ z^{2}$ hona chahiye (independent power-rule value: $\frac{d}{d z} z^{- 1} = - z^{- 2}$ ). Yeh check kyun? Agar $f^{'}$ ka hamara partial-derivative assembly is known formula se match kare, toh poora calculation corroborate ho jata hai. Assemble karo $f^{'} = u_{x} + i v_{x} = \frac{y ^{2} - x ^{2}}{D ^{2}} + i \frac{2 x y}{D ^{2}} = \frac{( y ^{2} - x ^{2} ) + i 2 x y}{D ^{2}}$ . Ab numerator: notice karo $- (x - i y)^{2} = - (x^{2} - y^{2} - 2 i x y) = (y^{2} - x^{2}) + 2 i x y$ — exactly hamara numerator hai, toh yeh $- \overset{z}{ˉ}^{2}$ ke equal hai. Aur $D = z \overset{z}{ˉ}$ toh $D^{2} = (z \overset{z}{ˉ})^{2} = z^{2} \overset{z}{ˉ}^{2}$ . Isliye $f^{'} = \frac{- z ˉ ^{2}}{z ^{2} z ˉ ^{2}} = \frac{- 1}{z ^{2}}$ ✅ ( $\overset{z}{ˉ}^{2}$ cleanly cancel ho jaata hai — koi sign trickery nahi, sirf numerator ko $- \overset{z}{ˉ}^{2}$ ke roop mein factor karna).

Ex 4 — Cell D: CR ek curve par, phir bhi analytic nahi

Worked example

$f (z) = z \overset{z}{ˉ} + z = ∣ z ∣^{2} + z$ . Forecast: $z$ entire hai lekin $∣ z ∣^{2}$ nowhere-analytic tha (parent Ex 4). Unka sum — CR kahan bachta hai?

Step 1. $∣ z ∣^{2} = x^{2} + y^{2}$ aur $z = x + i y$ , toh $u = x^{2} + y^{2} + x$ , $v = y$ . Yeh step kyun? Do pieces ke real aur imaginary parts alag-alag add karo.

Step 2. $u_{x} = 2 x + 1, u_{y} = 2 y, v_{x} = 0, v_{y} = 1$ . Yeh step kyun? CR $u_{x}$ ko $v_{y}$ se aur $u_{y}$ ko $- v_{x}$ se compare karta hai, isliye — har example ki tarah — koi bhi equation impose karne se pehle charon partials lay out karo.

Step 3. Pehli CR: $2 x + 1 = 1 \Rightarrow x = 0$ . Doosri CR: $u_{y} = 2 y$ vs $- v_{x} = 0 \Rightarrow y = 0$ . Yeh step kyun? Hum wo set locate karte hain jo dono satisfy kare.

Step 4. Dono sirf $(0, 0)$ par hold karte hain — ek point, open nahi ⇒ kahin bhi analytic nahi, lekin $z = 0$ par complex-differentiable hai. Yeh step kyun? Parent ki key subtlety reinforce karta hai: ek point par differentiability ≠ analyticity.

Verify: $z = 0$ par: $u_{x} = 1 = v_{y}$ ✅, $u_{y} = 0 = - v_{x}$ ✅ — exactly wahan differentiable. $z = 1$ par ( $x = 1$ ): $u_{x} = 3 \neq = v_{y} = 1$ ✅ fails, confirm karta hai ki yeh origin se door khatam ho jata hai. Yeh check kyun? Special set par ek point aur usse door ek point test karna dono halves of the conclusion confirm karta hai — CR $0$ par alive, baaki jagah dead.

Ex 5 — Cell E: harmonic conjugate build karo

Worked example Diya gaya hai

$u (x, y) = x^{3} - 3 x y^{2}$ , $v$ dhundho taaki $f = u + i v$ analytic ho, phir $f$ identify karo. Forecast: Aapne yeh $u$ Ex 1 mein dekha tha. Kya aap kaam karne se pehle poore $f$ ka guess kar sakte hain?

Step 1 — sanity: kya $u$ harmonic hai? $u_{xx} = 6 x$ , $u_{y y} = - 6 x$ , sum $= 0$ ✅. Yeh step kyun? Sirf harmonic $u$ ka ek conjugate ho sakta hai; parent ne prove kiya tha ki kisi bhi analytic function ka real part Laplace's equation satisfy karna chahiye. Dekho Harmonic functions and Laplace's equation.

Step 2 — CR se $v_{y}$ nikalo. $v_{y} = u_{x} = 3 x^{2} - 3 y^{2}$ . Yeh step kyun? Pehli CR equation humein known $u$ se $v$ ka ek partial recover karne deti hai.

Step 3 — $y$ mein integrate karo. $v = \int (3 x^{2} - 3 y^{2}) d y = 3 x^{2} y - y^{3} + g (x)$ , jahan $g (x)$ ek unknown "constant of integration" hai jo $x$ par depend kar sakta hai. Yeh step kyun? $v_{y}$ ko $y$ par integrate karna $v$ ko ek purely- $x$ function tak rebuild karta hai, kyunki $\partial / \partial y$ kuch bhi destroy kar deta hai jismein $y$ involve nahi hai.

Step 4 — doosri CR equation use karke $g$ pin karo. $v_{x} = 6 x y + g^{'} (x)$ , aur CR demand karta hai $v_{x} = - u_{y} = - (- 6 x y) = 6 x y$ . Toh $g^{'} (x) = 0 \Rightarrow g = C$ (ek genuine constant). Yeh step kyun? Doosri CR equation woh extra condition hai jo integration se bachi freedom ko remove karti hai.

Step 5 — assemble karo. $v = 3 x^{2} y - y^{3} + C$ , toh $f = (x^{3} - 3 x y^{2}) + i (3 x^{2} y - y^{3}) + i C = z^{3} + i C$ . Yeh step kyun? Ex 1 ka pattern recognize karo: yeh exactly $z^{3}$ hai (plus ek imaginary constant).

Verify: $f = z^{3} + i C$ entire hai (Ex 1). Iska real part $x^{3} - 3 x y^{2} = u$ hai ✅. Yeh check kyun? Finished $f$ se original $u$ reconstruct karna confirm karta hai ki conjugate bina error ke build hua. Conjugate additive constant $C$ tak unique hai — expected.

Ex 6 — Cell F: polar form, charon quadrants

Kuch functions — powers, roots, $lo g$ — polar coordinates $z = r (cos θ + i sin θ)$ mein bahut cleaner hote hain (dekho Complex numbers — polar form and Euler's formula). In ke liye hum polar Cauchy–Riemann equations use karte hain, jo multivariable chain rule se derive hoti hain (dekho Jacobian and the multivariable chain rule):

Neeche wali figure $1/ r$ ko concrete banati hai. Teal line vector $z = r e^{i θ}$ hai; plum arrow ek pure radial step hai (push $r$ by one unit — tip seedha bahar slide karti hai). Orange arc ek pure angular step of one radian hai: kyunki yeh radius $r$ par hota hai, tip ek $r$ length ka arc travel karti hai, $1$ nahi. Toh "per radian" change mein already $r$ extra travel ka factor bundled hai — precisely wahi $1/ r$ hai jo radial rate ( $\partial_{r}$ ) aur angular rate ( $\partial_{θ}$ ) ko same units mein laane ke liye divide karna hota hai. Picture se polar CR equations padho: $u$ ka radial-rate = $v$ ka (angular-rate) $1/ r$ se rescaled.

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

Worked example

$f (z) = lo g z = ln r + i θ$ . Yeh kahan analytic hai, aur kya cheez break karti hai? Forecast: $lo g$ mein zaroor koi trouble spot hoga — kahan, aur $z$ ka quadrant kyun matter karta hai?

Step 1. Read off karo $u = ln r$ , $v = θ$ . Yeh step kyun? Euler's formula directly $lo g z = ln r + i θ$ deta hai; polar naturally idhar fit baithta hai.

Step 2. Polar partials: $u_{r} = \frac{1}{r}, u_{θ} = 0, v_{r} = 0, v_{θ} = 1$ . Yeh step kyun? Polar CR equations in charon partials ke terms mein stated hain, isliye — exactly Cartesian examples ki tarah — do equations test karne se pehle hume table par har ingredient rakhna hoga.

Step 3. Polar CR check karo. $u_{r} = \frac{1}{r}$ aur $\frac{1}{r} v_{θ} = \frac{1}{r} \cdot 1 = \frac{1}{r}$ ✅. $v_{r} = 0$ aur $- \frac{1}{r} u_{θ} = 0$ ✅. Yeh step kyun? Dono polar CR equations har $r > 0$ ke liye hold karti hain.

Step 4 — quadrant subtlety. $θ$ tabhi single-valued hai jab hum plane cut karein. Principal branch $θ \in (- π, π]$ leta hai:

Quadrant I ( $x > 0, y > 0$ ): $θ = arctan (y / x) \in (0, \frac{π}{2})$ — smooth.
Quadrant II ( $x < 0, y > 0$ ): $θ \in (\frac{π}{2}, π)$ — smooth.
Quadrant III ( $x < 0, y < 0$ ): $θ \in (- π, - \frac{π}{2})$ — smooth.
Quadrant IV ( $x > 0, y < 0$ ): $θ \in (- \frac{π}{2}, 0)$ — smooth.
Negative real axis cross karna ( $y \to 0^{\pm}, x < 0$ ): $θ$ , $π$ se $- π$ tak jump karta hai. Discontinuity! Aur $r = 0$ (origin) undefined hai kyunki $ln 0 = - \infty$ . Yeh step kyun? CR pointwise hold karta hai, lekin analyticity ke liye values ki continuity chahiye; branch cut aur origin exactly wahan hain jahan continuity fail hoti hai. Har quadrant individually theek hai — failure sirf boundary line par hai.

Step 5. Conclusion: $lo g z$ , $C$ par minus the ray ${x \leq 0, y = 0}$ (negative real axis + origin) analytic hai.

Verify: Derivative $1/ z$ hona chahiye. Polar mein derivative formula hai $f^{'} = (cos θ - i sin θ) (u_{r} + i v_{r}) = e^{- i θ} \cdot \frac{1}{r} = \frac{1}{r e ^{i θ}} = \frac{1}{z}$ ✅. Yeh check kyun? $\frac{d}{d z} lo g z = 1/ z$ ek known real-analog result hai; isse match karna confirm karta hai ki polar partials correctly use hue.

Ex 7 — Cell G: real-world word problem

Worked example Ek ideal 2D fluid ka velocity potential

$u (x, y) = x^{2} - y^{2}$ hai. Physicists kehte hain ki flow "irrotational aur incompressible" exactly tabhi hai jab $u$ harmonic ho aur complex potential $f = u + i v$ analytic ho, jahan $v$ stream function hai (iski level curves flow lines hain). $v$ aur complex potential dhundho. Forecast: Aapne secretly yeh $u$ pehle dekha hai (yeh $Re z^{2}$ hai). Jawab guess karo.

Step 1 — harmonic check karo. $u_{xx} = 2$ , $u_{y y} = - 2$ , sum $= 0$ ✅. Physically: koi sources ya sinks nahi. Yeh step kyun? Incompressible + irrotational flow ⇔ harmonic potential; yeh physics gate hai. Harmonic functions and Laplace's equation se connect karta hai.

Step 2 — CR deta hai $v_{y} = u_{x} = 2 x$ . $y$ mein integrate karo: $v = 2 x y + g (x)$ . Yeh step kyun? Wahi conjugate-building machine jaise Ex 5 mein.

Step 3 — doosri CR. $v_{x} = 2 y + g^{'} (x)$ equal hona chahiye $- u_{y} = - (- 2 y) = 2 y$ , toh $g^{'} (x) = 0$ , $g = C$ . Yeh step kyun? Pehli CR equation ne unknown $x$ -only function $g (x)$ free chhod diya tha; doosri CR equation woh extra constraint hai jo $g^{'} (x)$ fix karti hai aur $v$ complete karti hai.

Step 4 — assemble karo. $v = 2 x y + C$ , aur $f = (x^{2} - y^{2}) + i 2 x y = z^{2}$ ( $C = 0$ choose karke). Yeh step kyun? Stream function analytic complex potential $f (z) = z^{2}$ complete karta hai — yeh corner ke around flow describe karta hai.

Verify: Velocity field hai $\overline{f^{'} (z)} = \overline{2 z} = 2 x - 2 i y$ , yaani velocity $(2 x, - 2 y)$ . Yeh check kyun? Do defining physical properties answer par hold honi chahiye, warna model galat hai. Divergence $= \partial_{x} (2 x) + \partial_{y} (- 2 y) = 2 + (- 2) = 0$ ✅ (incompressible — koi fluid create ya destroy nahi). Curl (2D scalar) $= \partial_{x} (- 2 y) - \partial_{y} (2 x) = 0 - 0 = 0$ ✅ (irrotational — koi local spinning nahi). Edge case: origin par velocity $(2 x, - 2 y)$ , $(0, 0)$ hai — ek stagnation point jahan flow momentarily ruk jaata hai, exactly "flow around a right-angle corner" picture ka corner; potential wahan perfectly analytic hai, toh kuch degenerate nahi hota. Units: potential carries units (velocity·length); iska gradient velocity return karta hai — dimensionally consistent.

Ex 8 — Cell H: exam-style parameter twist

Worked example Kin real constants

$a, b$ ke liye $f (z) = (x^{2} + a x y + b y^{2}) + i (x^{2} + 2 x y - y^{2})$ , $C$ par analytic hai? Forecast: Do unknowns, do CR equations — expect unique answer (ya contradiction). Guess karo ki yeh solvable hai ya nahi.

Step 1. Identify karo $u = x^{2} + a x y + b y^{2}$ , $v = x^{2} + 2 x y - y^{2}$ . Yeh step kyun? CR ek statement hai $u$ aur $v$ ke baare mein alag-alag, isliye kuch bhi differentiate karne se pehle hume padhna hoga ki real part kaunsi hai aur imaginary kaunsi.

Step 2. Partials: $u_{x} = 2 x + a y, u_{y} = a x + 2 b y, v_{x} = 2 x + 2 y, v_{y} = 2 x - 2 y$ . Yeh step kyun? Do CR equations $u_{x}$ ko $v_{y}$ se aur $u_{y}$ ko $- v_{x}$ se compare karti hain; hume charon partials table par chahiye (inme unknowns $a, b$ hain) koi bhi equation impose karne se pehle.

Step 3 — pehli CR: $u_{x} = v_{y}$ . $2 x + a y = 2 x - 2 y$ sabhi $x, y$ ke liye ⇒ coefficients match karo: $x$ -terms $2 = 2$ ✅, $y$ -terms $a = - 2$ . Yeh step kyun? $x, y$ mein ek identity coefficient-by-coefficient equality force karti hai.

Step 4 — doosri CR: $u_{y} = - v_{x}$ . $a x + 2 b y = - (2 x + 2 y) = - 2 x - 2 y$ . Match karo: $a = - 2$ ✅ (consistent!) aur $2 b = - 2 \Rightarrow b = - 1$ . Yeh step kyun? Doosri equation remaining unknown pin karti hai aur pehli ko cross-check karti hai — dono $a$ par agree karni chahiye, aur karti hain.

Step 5 — continuity verify karo. $a = - 2, b = - 1$ ke saath, sabhi partials linear hain ⇒ har jagah continuous hain ⇒ $f$ entire hai. Yeh step kyun? CR sirf necessary hai; sufficiency theorem ko ek open set par continuous partials chahiye, isliye $f$ analytic declare karne se pehle continuity confirm karni hogi — wahi guard jo parent ke mistake callout ne insist kiya tha.

Verify: $a = - 2, b = - 1$ ke saath: $u = x^{2} - 2 x y - y^{2}$ , $v = x^{2} + 2 x y - y^{2}$ . Group karo $u + i v = (1 + i) (x^{2} - y^{2}) + (- 2 + 2 i) x y$ . $(1 + i) z^{2} = (1 + i) (x^{2} - y^{2} + 2 i x y) = (1 + i) (x^{2} - y^{2}) + 2 i (1 + i) x y = (1 + i) (x^{2} - y^{2}) + (2 i - 2) x y$ se compare karo — match karta hai, toh $f = (1 + i) z^{2}$ , manifestly entire. Yeh check kyun? Agar answer values $f$ ko sirf $z$ ka ek clean function banate hain ( $\overset{z}{ˉ}$ nahi), toh analyticity free mein guaranteed hai — ek strong confirmation ki $a = - 2, b = - 1$ sahi hai.

Recall Aapke liye kaun sa cell sabse mushkil tha?

Matrix cover karo aur sirf statement se har row ka ek example re-derive karo. Cell D vs C — kya fark hai? ::: C ek isolated point par fail hota hai apne domain se bahar (ek pole, baaki jagah analytic); D apne domain ke andar ek measure-zero set par CR satisfy karta hai (kabhi analytic nahi). Cell F — $lo g z$ negative real axis par kyun break hua? ::: $θ$ , branch cut ke across $2 π$ jump karta hai, isliye values wahan discontinuous hain chahe polar CR pointwise hold kare.

Active recall

Harmonic

u

diya gaya hai,

v

recover karne ke liye do CR steps kya hain?

v_{y} = u_{x}

y

mein integrate karo (gives

v

up to

g (x)

), phir

g^{'} (x)

solve karne ke liye

v_{x} = - u_{y}

use karo.

Polar Cauchy–Riemann equations state karo.

u_{r} = \frac{1}{r} v_{θ}

aur

v_{r} = - \frac{1}{r} u_{θ}

1/ z

kyon

C ∖ {0}

par analytic hai lekin

∣ z ∣^{2}

nowhere analytic hai?

1/ z

poore punctured plane par CR satisfy karta hai (ek open set);

∣ z ∣^{2}

sirf single non-open point

0

par CR satisfy karta hai.