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

We only use notation the parent already built: $z = x + i y$ , $f (z) = u (x, y) + i v (x, y)$ , the partial derivatives $u_{x} = \partial u / \partial x$ (rate of change of $u$ as $x$ moves, $y$ frozen), and the two CR equations. If any of that feels shaky, reread the parent note first.

The scenario matrix

Every problem this topic throws lands in exactly one of these cells. Each row is a class of behaviour; the last column names the example that nails it.

Cell	What makes it distinct	Example
A. Analytic everywhere (entire)	CR holds at every point; partials continuous	Ex 1: $z^{3}$
B. Nowhere analytic	CR fails at every point (orientation-reversing)	Ex 2: $f = \overset{z}{ˉ}^{2}$
C. Analytic except a pole	CR holds off one bad point where $f$ blows up	Ex 3: $1/ z$
D. CR on a curve only (degenerate)	CR holds on a line/point, no open set ⇒ not analytic	Ex 4: $z \overset{z}{ˉ} + z$
E. Build the conjugate	Given harmonic $u$ , construct $v$ so $f$ is analytic	Ex 5: $u = x^{3} - 3 x y^{2}$
F. Polar / non-Cartesian input	Function given in $r, θ$ — need polar CR	Ex 6: $lo g z$
G. Real-world word problem	Physics dressed as analyticity (flow / potential)	Ex 7: fluid potential
H. Exam-style twist	"For which $a, b$ …" — solve for a parameter	Ex 8: $f = x^{2} + a x y + b y^{2} + i (\dots)$

The signs and quadrants (the "every quadrant" demand) show up inside Ex 6, where $θ$ ranges over all four quadrants and the branch of $lo g$ must be chosen carefully — that is where sign-of- $x$ , sign-of- $y$ logic lives for this topic.

Ex 1 — Cell A: analytic everywhere

Worked example

$f (z) = z^{3}$ . Is it analytic? Where? Forecast: Guess before reading — polynomial in $z$ , so… everywhere? Nowhere? One point? Commit to an answer.

Step 1. Expand into real and imaginary parts. $z^{3} = (x + i y)^{3} = x^{3} + 3 x^{2} (i y) + 3 x (i y)^{2} + (i y)^{3} = x^{3} - 3 x y^{2} + i (3 x^{2} y - y^{3})$ . So $u = x^{3} - 3 x y^{2}$ , $v = 3 x^{2} y - y^{3}$ . Why this step? CR is stated in terms of $u$ and $v$ , so we must physically separate them before testing anything.

Step 2. Compute all four partials. $u_{x} = 3 x^{2} - 3 y^{2}, u_{y} = - 6 x y, v_{x} = 6 x y, v_{y} = 3 x^{2} - 3 y^{2}$ . Why this step? CR compares $u_{x}$ with $v_{y}$ and $u_{y}$ with $- v_{x}$ ; we need every ingredient on the table.

Step 3. Check both CR equations. $u_{x} = 3 x^{2} - 3 y^{2} = v_{y}$ ✅. $u_{y} = - 6 x y$ and $- v_{x} = - 6 x y$ ✅. Why this step? Both must hold; one alone is not enough.

Step 4. Continuity. All four partials are polynomials ⇒ continuous on all of $C$ . By the sufficiency theorem, $f$ is entire. Why this step? CR + continuous partials on an open set = analytic. Skipping continuity is the classic error the parent warned about.

Verify: The complex derivative should equal $3 z^{2}$ . Compute $f^{'} = u_{x} + i v_{x} = (3 x^{2} - 3 y^{2}) + i 6 x y = 3 (x^{2} - y^{2} + 2 i x y) = 3 (x + i y)^{2} = 3 z^{2}$ ✅. Why this check? An independent formula ( $\frac{d}{d z} z^{3} = 3 z^{2}$ from the power rule) must agree with the value we got by gluing partials together — agreement means no algebra slipped. Units-style sanity: cubing a length-2 vector triples its "spin rate" — consistent with the $3$ .

Ex 2 — Cell B: nowhere analytic

Worked example

$f (z) = \overset{z}{ˉ}^{2} = (x - i y)^{2}$ . Forecast: It's a square, and $z^{2}$ was entire. Does the bar ruin it everywhere, or just on a curve?

Step 1. Expand. $\overset{z}{ˉ}^{2} = (x - i y)^{2} = x^{2} - y^{2} - 2 i x y$ . So $u = x^{2} - y^{2}$ , $v = - 2 x y$ . Why this step? Same reason as always — CR needs $u, v$ split out.

Step 2. Partials: $u_{x} = 2 x, u_{y} = - 2 y, v_{x} = - 2 y, v_{y} = - 2 x$ . Why this step? The two CR equations compare $u_{x}$ with $v_{y}$ and $u_{y}$ with $- v_{x}$ , so we must have all four partials in hand before we can test either equation.

Step 3. First CR: $u_{x} = 2 x$ vs $v_{y} = - 2 x$ . Equal only when $2 x = - 2 x \Rightarrow x = 0$ . Second CR: $u_{y} = - 2 y$ vs $- v_{x} = 2 y$ . Equal only when $y = 0$ . Why this step? We're hunting the set where BOTH hold.

Step 4. Both hold only at $x = 0$ and $y = 0$ , i.e. the single point $z = 0$ . A single point is not an open set ⇒ nowhere analytic. Why this step? Analyticity requires an open neighborhood; one point can never provide it.

Verify: Geometrically $\overset{z}{ˉ}^{2}$ reflects (from the bar) then squares. The reflection reverses orientation, breaking the "spin-and-zoom only" handshake — matches "nowhere analytic." Why this check? A geometric argument that agrees with the algebra guards against a sign slip in the partials. Numeric spot-check at $z = 1$ ( $x = 1, y = 0$ ): $u_{x} = 2, v_{y} = - 2$ , unequal ✅ (CR genuinely fails away from origin).

Ex 3 — Cell C: analytic except one pole

Worked example

$f (z) = \frac{1}{z}$ . Where is it analytic? Forecast: Division by $z$ — trouble at $z = 0$ , but what about everywhere else?

Step 1. Rationalise: $\frac{1}{z} = \frac{z ˉ}{z z ˉ} = \frac{x - i y}{x ^{2} + y ^{2}}$ . So $u = \frac{x}{x ^{2} + y ^{2}}$ , $v = \frac{- y}{x ^{2} + y ^{2}}$ , valid wherever $x^{2} + y^{2} \neq = 0$ , i.e. $z \neq = 0$ . Why this step? We can only test CR where $u, v$ are actually defined; the domain excludes $0$ from the start.

Step 2. Let $D = x^{2} + y^{2}$ . Using the quotient rule: $u_{x} = \frac{D - x \cdot 2 x}{D ^{2}} = \frac{y ^{2} - x ^{2}}{D ^{2}}, v_{y} = \frac{- D - ( - y ) \cdot 2 y}{D ^{2}} = \frac{y ^{2} - x ^{2}}{D ^{2}}$ . So $u_{x} = v_{y}$ ✅. Why this step? Direct check of the first CR equation on the punctured plane.

Step 3. $u_{y} = \frac{- x \cdot 2 y}{D ^{2}} = \frac{- 2 x y}{D ^{2}}$ , and $v_{x} = \frac{- ( - y ) \cdot 2 x}{D ^{2}} = \frac{2 x y}{D ^{2}}$ , so $- v_{x} = \frac{- 2 x y}{D ^{2}} = u_{y}$ ✅. Why this step? The second CR equation must be checked too; passing only the first is exactly the half-check the parent's mistake callout warns against.

Step 4. Partials are rational functions, continuous everywhere except $z = 0$ . Hence $f$ is analytic on $C ∖ {0}$ ; the point $0$ is a pole, not part of the domain. Why this step? Continuity fails only where the denominator vanishes; that lone point is excluded, so the open-set requirement is met everywhere else.

Verify: $f^{'}$ should be $- 1/ z^{2}$ (independent power-rule value: $\frac{d}{d z} z^{- 1} = - z^{- 2}$ ). Why this check? If our partial-derivative assembly of $f^{'}$ matches this known formula, the whole calculation is corroborated. Assemble $f^{'} = u_{x} + i v_{x} = \frac{y ^{2} - x ^{2}}{D ^{2}} + i \frac{2 x y}{D ^{2}} = \frac{( y ^{2} - x ^{2} ) + i 2 x y}{D ^{2}}$ . Now the numerator: notice $- (x - i y)^{2} = - (x^{2} - y^{2} - 2 i x y) = (y^{2} - x^{2}) + 2 i x y$ — exactly our numerator, so it equals $- \overset{z}{ˉ}^{2}$ . And $D = z \overset{z}{ˉ}$ so $D^{2} = (z \overset{z}{ˉ})^{2} = z^{2} \overset{z}{ˉ}^{2}$ . Therefore $f^{'} = \frac{- z ˉ ^{2}}{z ^{2} z ˉ ^{2}} = \frac{- 1}{z ^{2}}$ ✅ (the $\overset{z}{ˉ}^{2}$ cancels cleanly — no sign trickery, just factoring the numerator as $- \overset{z}{ˉ}^{2}$ ).

Ex 4 — Cell D: CR on a curve, still not analytic

Worked example

$f (z) = z \overset{z}{ˉ} + z = ∣ z ∣^{2} + z$ . Forecast: $z$ is entire but $∣ z ∣^{2}$ was nowhere-analytic (parent Ex 4). Their sum — where does CR survive?

Step 1. $∣ z ∣^{2} = x^{2} + y^{2}$ and $z = x + i y$ , so $u = x^{2} + y^{2} + x$ , $v = y$ . Why this step? Add the two pieces' real and imaginary parts separately.

Step 2. $u_{x} = 2 x + 1, u_{y} = 2 y, v_{x} = 0, v_{y} = 1$ . Why this step? CR compares $u_{x}$ with $v_{y}$ and $u_{y}$ with $- v_{x}$ , so — as in every example — we lay out all four partials before imposing either equation.

Step 3. First CR: $2 x + 1 = 1 \Rightarrow x = 0$ . Second CR: $u_{y} = 2 y$ vs $- v_{x} = 0 \Rightarrow y = 0$ . Why this step? We locate the set satisfying both.

Step 4. Both hold only at $(0, 0)$ — one point, not open ⇒ not analytic anywhere, though complex-differentiable at $z = 0$ . Why this step? Reinforces the parent's key subtlety: differentiability at a point ≠ analyticity.

Verify: At $z = 0$ : $u_{x} = 1 = v_{y}$ ✅, $u_{y} = 0 = - v_{x}$ ✅ — differentiable exactly there. At $z = 1$ ( $x = 1$ ): $u_{x} = 3 \neq = v_{y} = 1$ ✅ fails, confirming it dies off the origin. Why this check? Testing one point on the special set and one point off it confirms both halves of the conclusion — CR alive at $0$ , dead elsewhere.

Ex 5 — Cell E: build the harmonic conjugate

Worked example Given

$u (x, y) = x^{3} - 3 x y^{2}$ , find $v$ so that $f = u + i v$ is analytic, then identify $f$ . Forecast: You saw this $u$ in Ex 1. Can you guess the whole $f$ before doing the work?

Step 1 — sanity: is $u$ harmonic? $u_{xx} = 6 x$ , $u_{y y} = - 6 x$ , sum $= 0$ ✅. Why this step? Only harmonic $u$ can have a conjugate; the parent proved the real part of any analytic function must satisfy Laplace's equation. See Harmonic functions and Laplace's equation.

Step 2 — use CR to get $v_{y}$ . $v_{y} = u_{x} = 3 x^{2} - 3 y^{2}$ . Why this step? The first CR equation lets us recover a partial of $v$ from the known $u$ .

Step 3 — integrate in $y$ . $v = \int (3 x^{2} - 3 y^{2}) d y = 3 x^{2} y - y^{3} + g (x)$ , where $g (x)$ is an unknown "constant of integration" that may depend on $x$ . Why this step? Integrating $v_{y}$ over $y$ rebuilds $v$ up to a purely- $x$ function, because $\partial / \partial y$ kills anything not involving $y$ .

Step 4 — pin down $g$ using the second CR equation. $v_{x} = 6 x y + g^{'} (x)$ , and CR demands $v_{x} = - u_{y} = - (- 6 x y) = 6 x y$ . So $g^{'} (x) = 0 \Rightarrow g = C$ (a genuine constant). Why this step? The second CR equation is the extra condition that removes the freedom left by integration.

Step 5 — assemble. $v = 3 x^{2} y - y^{3} + C$ , so $f = (x^{3} - 3 x y^{2}) + i (3 x^{2} y - y^{3}) + i C = z^{3} + i C$ . Why this step? Recognise the pattern from Ex 1: this is exactly $z^{3}$ (plus an imaginary constant).

Verify: $f = z^{3} + i C$ is entire (Ex 1). Its real part is $x^{3} - 3 x y^{2} = u$ ✅. Why this check? Reconstructing the original $u$ from the finished $f$ confirms the conjugate was built without error. Conjugate is unique up to the additive constant $C$ — expected.

Ex 6 — Cell F: polar form, all four quadrants

Some functions — powers, roots, $lo g$ — are far cleaner in polar coordinates $z = r (cos θ + i sin θ)$ (see Complex numbers — polar form and Euler's formula). For these we use the polar Cauchy–Riemann equations, derived by the multivariable chain rule (see Jacobian and the multivariable chain rule):

The figure below makes the $1/ r$ concrete. The teal line is the vector $z = r e^{i θ}$ ; the plum arrow is a pure radial step (push $r$ by one unit — the tip slides straight outward). The orange arc is a pure angular step of one radian: because it happens at radius $r$ , the tip travels an arc of length $r$ , not $1$ . So a change "per radian" already bundles in a factor of $r$ of extra travel — that is precisely the $1/ r$ we must divide out to make the radial rate ( $\partial_{r}$ ) and the angular rate ( $\partial_{θ}$ ) speak the same units. Read the polar CR equations off the picture: radial-rate of $u$ = (angular-rate of $v$ ) rescaled by $1/ r$ .

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

Worked example

$f (z) = lo g z = ln r + i θ$ . Where is it analytic, and what breaks? Forecast: $lo g$ has to have some trouble spot — where, and why does the quadrant of $z$ matter?

Step 1. Read off $u = ln r$ , $v = θ$ . Why this step? Euler's formula gives $lo g z = ln r + i θ$ directly; polar is the natural home.

Step 2. Polar partials: $u_{r} = \frac{1}{r}, u_{θ} = 0, v_{r} = 0, v_{θ} = 1$ . Why this step? The polar CR equations are stated in terms of these four partials, so — exactly as in the Cartesian examples — we must lay every ingredient on the table before we can test the two equations.

Step 3. Check polar CR. $u_{r} = \frac{1}{r}$ and $\frac{1}{r} v_{θ} = \frac{1}{r} \cdot 1 = \frac{1}{r}$ ✅. $v_{r} = 0$ and $- \frac{1}{r} u_{θ} = 0$ ✅. Why this step? Both polar CR equations hold for every $r > 0$ .

Step 4 — the quadrant subtlety. $θ$ is only single-valued if we cut the plane. The principal branch takes $θ \in (- π, π]$ :

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.
Crossing the negative real axis ( $y \to 0^{\pm}, x < 0$ ): $θ$ jumps from $π$ to $- π$ . Discontinuity! And $r = 0$ (the origin) is undefined since $ln 0 = - \infty$ . Why this step? CR holds pointwise, but analyticity needs continuity of the values; the branch cut and the origin are exactly where continuity fails. Every quadrant is individually fine — the failure is only on the boundary line.

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

Verify: Derivative should be $1/ z$ . In polar the derivative formula is $f^{'} = (cos θ - i sin θ) (u_{r} + i v_{r}) = e^{- i θ} \cdot \frac{1}{r} = \frac{1}{r e ^{i θ}} = \frac{1}{z}$ ✅. Why this check? $\frac{d}{d z} lo g z = 1/ z$ is the known real-analog result; matching it confirms the polar partials were used correctly.

Ex 7 — Cell G: real-world word problem

Worked example An ideal 2D fluid has velocity potential

$u (x, y) = x^{2} - y^{2}$ . Physicists say the flow is "irrotational and incompressible" exactly when $u$ is harmonic and the complex potential $f = u + i v$ is analytic, where $v$ is the stream function (its level curves are the flow lines). Find $v$ and the complex potential. Forecast: You've secretly seen this $u$ before (it's $Re z^{2}$ ). Guess the answer.

Step 1 — check harmonic. $u_{xx} = 2$ , $u_{y y} = - 2$ , sum $= 0$ ✅. Physically: no sources or sinks. Why this step? Incompressible + irrotational flow ⇔ harmonic potential; this is the physics gate. Connects to Harmonic functions and Laplace's equation.

Step 2 — CR gives $v_{y} = u_{x} = 2 x$ . Integrate in $y$ : $v = 2 x y + g (x)$ . Why this step? Same conjugate-building machine as Ex 5.

Step 3 — second CR. $v_{x} = 2 y + g^{'} (x)$ must equal $- u_{y} = - (- 2 y) = 2 y$ , so $g^{'} (x) = 0$ , $g = C$ . Why this step? The first CR equation left the unknown $x$ -only function $g (x)$ free; the second CR equation is the extra constraint that fixes $g^{'} (x)$ and completes $v$ .

Step 4 — assemble. $v = 2 x y + C$ , and $f = (x^{2} - y^{2}) + i 2 x y = z^{2}$ (choosing $C = 0$ ). Why this step? The stream function completes the analytic complex potential $f (z) = z^{2}$ — this describes flow around a corner.

Verify: The velocity field is $\overline{f^{'} (z)} = \overline{2 z} = 2 x - 2 i y$ , i.e. velocity $(2 x, - 2 y)$ . Why this check? The two defining physical properties must hold on the answer, or the model is wrong. Divergence $= \partial_{x} (2 x) + \partial_{y} (- 2 y) = 2 + (- 2) = 0$ ✅ (incompressible — no fluid created or destroyed). Curl (2D scalar) $= \partial_{x} (- 2 y) - \partial_{y} (2 x) = 0 - 0 = 0$ ✅ (irrotational — no local spinning). Edge case: at the origin the velocity $(2 x, - 2 y)$ is $(0, 0)$ — a stagnation point where the flow momentarily stops, exactly the corner of the "flow around a right-angle corner" picture; the potential is perfectly analytic there, so nothing degenerates. Units: potential carries units (velocity·length); its gradient returns a velocity — dimensionally consistent.

Ex 8 — Cell H: exam-style parameter twist

Worked example For which real constants

$a, b$ is $f (z) = (x^{2} + a x y + b y^{2}) + i (x^{2} + 2 x y - y^{2})$ analytic on $C$ ? Forecast: Two unknowns, two CR equations — expect a unique answer (or a contradiction). Guess whether it's solvable.

Step 1. Identify $u = x^{2} + a x y + b y^{2}$ , $v = x^{2} + 2 x y - y^{2}$ . Why this step? CR is a statement about $u$ and $v$ separately, so before differentiating anything we must read off which part is real and which is imaginary.

Step 2. Partials: $u_{x} = 2 x + a y, u_{y} = a x + 2 b y, v_{x} = 2 x + 2 y, v_{y} = 2 x - 2 y$ . Why this step? The two CR equations compare $u_{x}$ with $v_{y}$ and $u_{y}$ with $- v_{x}$ ; we need all four partials on the table (they carry the unknowns $a, b$ ) before we can impose those equations.

Step 3 — first CR: $u_{x} = v_{y}$ . $2 x + a y = 2 x - 2 y$ for all $x, y$ ⇒ match coefficients: $x$ -terms $2 = 2$ ✅, $y$ -terms $a = - 2$ . Why this step? An identity in $x, y$ forces coefficient-by-coefficient equality.

Step 4 — second CR: $u_{y} = - v_{x}$ . $a x + 2 b y = - (2 x + 2 y) = - 2 x - 2 y$ . Match: $a = - 2$ ✅ (consistent!) and $2 b = - 2 \Rightarrow b = - 1$ . Why this step? The second equation pins the remaining unknown and cross-checks the first — both must agree on $a$ , and they do.

Step 5 — verify continuity. With $a = - 2, b = - 1$ , all partials are linear ⇒ continuous everywhere ⇒ $f$ entire. Why this step? CR alone is only necessary; the sufficiency theorem needs continuous partials on an open set, so we must confirm continuity before declaring $f$ analytic — the same guard the parent's mistake callout insists on.

Verify: With $a = - 2, b = - 1$ : $u = x^{2} - 2 x y - y^{2}$ , $v = x^{2} + 2 x y - y^{2}$ . Group $u + i v = (1 + i) (x^{2} - y^{2}) + (- 2 + 2 i) x y$ . Check against $(1 + i) z^{2} = (1 + i) (x^{2} - y^{2} + 2 i x y) = (1 + i) (x^{2} - y^{2}) + 2 i (1 + i) x y = (1 + i) (x^{2} - y^{2}) + (2 i - 2) x y$ — matches, so $f = (1 + i) z^{2}$ , manifestly entire. Why this check? If the answer values collapse $f$ into a clean function of $z$ alone (no $\overset{z}{ˉ}$ ), analyticity is guaranteed for free — a strong confirmation that $a = - 2, b = - 1$ is right.

Recall Which cell was hardest for you?

Cover the matrix and re-derive one example per row from the statement alone. Cell D vs C — what's the difference? ::: C fails at an isolated point outside its domain (a pole, still analytic elsewhere); D satisfies CR only on a measure-zero set inside its domain (never analytic). Cell F — why did $lo g z$ break on the negative real axis? ::: $θ$ jumps by $2 π$ across the branch cut, so the values are discontinuous there even though polar CR holds pointwise.

Active recall

Given harmonic

u

, what two CR steps recover

v

Integrate

v_{y} = u_{x}

y

(gives

v

up to

g (x)

), then use

v_{x} = - u_{y}

to solve for

g^{'} (x)

State the polar Cauchy–Riemann equations.

u_{r} = \frac{1}{r} v_{θ}

and

v_{r} = - \frac{1}{r} u_{θ}

Why is

1/ z

analytic on

C ∖ {0}

but

∣ z ∣^{2}

nowhere analytic?

1/ z

satisfies CR on the whole punctured plane (an open set);

∣ z ∣^{2}

satisfies CR only at the single non-open point

0