Complex analysis — analytic functions, Cauchy-Riemann equations: Exercises

Here $u_{x}$ means $\partial u / \partial x$ , "the rate of change of $u$ as only $x$ moves," and likewise for the others. Every symbol below is used exactly as the parent defined it.

Level 1 — Recognition

L1·Q1. Read off $u$ and $v$

For $f (z) = (2 - 3 i) z$ , write $f$ in the form $u (x, y) + i v (x, y)$ .

Recall Solution — L1·Q1

Substitute $z = x + i y$ : $f = (2 - 3 i) (x + i y) = 2 x + 2 i y - 3 i x - 3 i^{2} y = (2 x + 3 y) + i (2 y - 3 x) .$ since $i^{2} = - 1$ . So $u = 2 x + 3 y, v = - 3 x + 2 y .$ WHY the split works: any complex number splits uniquely into (real part) $+ i$ (imaginary part); we just collected the terms with no $i$ into $u$ and the terms multiplying $i$ into $v$ .

L1·Q2. Does this pair even look like it could satisfy CR?

Given $u = x^{2} - y^{2}$ and $v = 2 x y$ , compute all four partials $u_{x}, u_{y}, v_{x}, v_{y}$ (don't check CR yet — just list them).

Recall Solution — L1·Q2

Differentiate treating the other variable as a constant:

$u_{x} = \frac{\partial}{\partial x} (x^{2} - y^{2}) = 2 x$ (the $- y^{2}$ is constant in $x$ ),
$u_{y} = \frac{\partial}{\partial y} (x^{2} - y^{2}) = - 2 y$ ,
$v_{x} = \frac{\partial}{\partial x} (2 x y) = 2 y$ ,
$v_{y} = \frac{\partial}{\partial y} (2 x y) = 2 x$ . WHAT this is: the four entries of the Jacobian $(u_{x} v_{x} u_{y} v_{y}) = (2 x 2 y - 2 y 2 x)$ — notice the $(a b - b a)$ "rotation+scale" shape flagged in the parent.

Level 2 — Application

L2·Q1. Full analyticity test

Is $f (z) = e^{x} cos y - i e^{x} sin y$ analytic anywhere? If so, find $f^{'} (z)$ in terms of $z$ .

Recall Solution — L2·Q1

Here $u = e^{x} cos y$ , $v = - e^{x} sin y$ . $u_{x} = e^{x} cos y, u_{y} = - e^{x} sin y, v_{x} = - e^{x} sin y, v_{y} = - e^{x} cos y .$ CR check 1: $u_{x} = e^{x} cos y$ but $v_{y} = - e^{x} cos y$ . These are equal only where $e^{x} cos y = 0$ , i.e. $cos y = 0$ (since $e^{x} > 0$ always). That is a set of isolated lines, not an open set. Conclusion: CR fails on every open region, so $f$ is nowhere analytic. (In fact $f (z) = e^{x} (cos y - i sin y) = e^{x} e^{- i y} = e^{\overset{z}{ˉ}}$ — a function of $\overset{z}{ˉ}$ , which the parent warned can't be analytic.)

L2·Q2. Find the constant that makes it analytic

For which real constant $a$ is $u = x^{2} + a y^{2}$ the real part of an analytic function? Then find a valid $v$ .

Recall Solution — L2·Q2

An analytic $u$ must be harmonic: $u_{xx} + u_{y y} = 0$ (parent's Laplace result). $u_{xx} = 2, u_{y y} = 2 a \Rightarrow 2 + 2 a = 0 \Rightarrow a = - 1.$ So $u = x^{2} - y^{2}$ . Recover $v$ from CR:

$v_{y} = u_{x} = 2 x \Rightarrow v = 2 x y + g (x)$ (integrate in $y$ ; the "constant" can depend on $x$ ).
$v_{x} = - u_{y}$ . Now $u_{y} = - 2 y$ so $- u_{y} = 2 y$ , and $v_{x} = 2 y + g^{'} (x)$ . Matching: $2 y + g^{'} (x) = 2 y \Rightarrow g^{'} (x) = 0 \Rightarrow g = const$ . $a = - 1, v = 2 x y + C .$ This rebuilds $f (z) = z^{2}$ (up to the imaginary constant $i C$ ), matching Example 1 of the parent. See Harmonic functions and Laplace's equation for why "harmonic" is the gatekeeper.

Level 3 — Analysis

L3·Q1. CR at a point but not analytic

Show $f (z) = z Re (z) = z x$ satisfies CR only on a line, and state where (if anywhere) it is differentiable vs analytic.

Recall Solution — L3·Q1

$f = (x + i y) x = x^{2} + i x y$ , so $u = x^{2}$ , $v = x y$ . $u_{x} = 2 x, v_{y} = x, u_{y} = 0, v_{x} = y .$

CR-1: $u_{x} = v_{y} \Rightarrow 2 x = x \Rightarrow x = 0$ .
CR-2: $u_{y} = - v_{x} \Rightarrow 0 = - y \Rightarrow y = 0$ . Both hold only at the single point $z = 0$ (the intersection $x = 0, y = 0$ ). A single point is not open, so $f$ is (at most) complex-differentiable at $0$ and nowhere analytic. The partials are continuous, so differentiability does hold at that one point, with $f^{'} (0) = u_{x} + i v_{x}_{0} = 0$ .

L3·Q2. Where does a "quadratic-with-conjugate" break?

Let $f (z) = z^{2} + \overset{z}{ˉ}$ . Find all points where CR holds. Is $f$ analytic anywhere?

Recall Solution — L3·Q2

$z^{2} = x^{2} - y^{2} + 2 i x y$ and $\overset{z}{ˉ} = x - i y$ , so $u = x^{2} - y^{2} + x, v = 2 x y - y .$ $u_{x} = 2 x + 1, v_{y} = 2 x - 1, u_{y} = - 2 y, v_{x} = 2 y .$

CR-1: $2 x + 1 = 2 x - 1 \Rightarrow 1 = - 1$ , impossible.
CR-2: $- 2 y = - 2 y$ , always true. CR-1 fails at every point, so $f$ is nowhere analytic. The offending piece is exactly the $\overset{z}{ˉ}$ : it single-handedly shifts $u_{x}$ and $v_{y}$ apart by the constant $2$ , and no choice of $x, y$ can heal a constant mismatch.

Level 4 — Synthesis

L4·Q1. Build the analytic function from a harmonic part

Given the harmonic function $u = e^{x} cos y$ , construct the analytic $f = u + i v$ (find $v$ ), and identify $f (z)$ in closed form. Which vault tool guaranteed a $v$ exists?

Recall Solution — L4·Q1

First confirm harmonic: $u_{xx} = e^{x} cos y$ , $u_{y y} = - e^{x} cos y$ , sum $= 0$ ✅ — so a harmonic conjugate $v$ is guaranteed to exist (locally), the fact from Harmonic functions and Laplace's equation. Use CR to recover $v$ :

$v_{y} = u_{x} = e^{x} cos y \Rightarrow v = e^{x} sin y + g (x)$ (integrate in $y$ ).
$v_{x} = - u_{y}$ . Compute $u_{y} = - e^{x} sin y$ , so $- u_{y} = e^{x} sin y$ . Also $v_{x} = e^{x} sin y + g^{'} (x)$ . Match: $g^{'} (x) = 0$ . $v = e^{x} sin y + C .$ Then $f = e^{x} cos y + i e^{x} sin y = e^{x} (cos y + i sin y) = e^{x} e^{i y} = e^{x + i y} = e^{z}$ (taking $C = 0$ ), using Euler's formula from Complex numbers — polar form and Euler's formula. This reconstructs Example 3 of the parent.

L4·Q2. CR forces rigidity

Suppose $f$ is analytic on an open connected region and its imaginary part $v$ is constant there. Prove $f$ is constant.

Recall Solution — L4·Q2

$v$ constant means $v_{x} = 0$ and $v_{y} = 0$ everywhere in the region. Apply CR:

$u_{x} = v_{y} = 0$ ,
$u_{y} = - v_{x} = 0$ . So all four partials of $u$ vanish, hence $u$ is locally constant; on a connected region "all partials zero" forces $u$ to be a single constant. Therefore $f = u + i v$ is constant. The moral: you cannot fix the imaginary part and freely wiggle the real part — CR chains them together. This rigidity is what makes analytic functions so powerful in Contour integration and Cauchy's integral theorem.

Level 5 — Mastery

L5·Q1. The Jacobian view — prove CR ⟺ "rotation+scale"

Let $F (x, y) = (u, v)$ be real-differentiable at a point with Jacobian $J = (u_{x} v_{x} u_{y} v_{y})$ . Prove that $J$ acts as multiplication by a single complex number $a + ib$ if and only if the CR equations hold, and that then $∣ f^{'} ∣^{2} = det J$ .

Recall Solution — L5·Q1

Setup. Multiplying a point $x + i y$ by the fixed complex number $a + ib$ gives $(a + ib) (x + i y) = (a x - b y) + i (b x + a y),$ so as a real-linear map on $(x, y)$ this multiplication is the matrix $M (a, b) = (a b - b a) .$ This is a rotation-by- $θ$ scaled by $r$ , where $a + ib = r e^{i θ}$ (see Complex numbers — polar form and Euler's formula).

(⇐) CR ⇒ special form. If $u_{x} = v_{y}$ and $u_{y} = - v_{x}$ , set $a = u_{x} = v_{y}$ and $b = v_{x} = - u_{y}$ . Then $J = (u_{x} v_{x} u_{y} v_{y}) = (a b - b a) = M (a, b) .$ So $J$ is exactly multiplication by $a + ib$ , and $f^{'} = u_{x} + i v_{x} = a + ib$ .

(⇒) Special form ⇒ CR. Conversely if $J = M (a, b)$ for some $a, b$ , then reading entries: $u_{x} = a = v_{y}$ (top-left = bottom-right) and $u_{y} = - b = - v_{x}$ (top-right = $-$ bottom-left). Those are precisely the two CR equations.

Determinant. For the special form, $det J = a \cdot a - (- b) \cdot b = a^{2} + b^{2} = ∣ a + ib ∣^{2} = ∣ f^{'} ∣^{2} .$ WHAT this means geometrically (figure below): the map scales every little area by $∣ f^{'} ∣^{2}$ and rotates by $ar g f^{'}$ , uniformly in all directions — that is the "spin-and-zoom, no shear, no flip" of the parent's rubber-sheet story, and the reason analytic maps are conformal (Conformal mappings). The Jacobian and the multivariable chain rule is the machinery that turns local stretching into this matrix.

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

L5·Q2. A discontinuity trap (CR holds at 0, yet not differentiable)

Consider $f (z) = ⎩ ⎨ ⎧ \frac{z ˉ ^{2}}{z}, 0, z \neq = 0, z = 0.$ Show that the CR equations hold at $z = 0$ , yet $f$ is not complex-differentiable at $0$ . Explain which hypothesis of the sufficiency theorem fails.

Recall Solution — L5·Q2

Step 1 — CR at $0$ via partial derivatives. Along the real axis ( $y = 0$ , $z = x$ ): $\overset{z}{ˉ} = x$ , so $f (x) = x^{2} / x = x$ , giving $u (x, 0) = x, v (x, 0) = 0$ . Hence $u_{x} (0, 0) = 1, v_{x} (0, 0) = 0$ . Along the imaginary axis ( $x = 0$ , $z = i y$ ): $\overset{z}{ˉ} = - i y$ , $\overset{z}{ˉ}^{2} = - y^{2}$ , and $\overset{z}{ˉ}^{2} / z = (- y^{2}) / (i y) = - y / i = i y$ . So $u (0, y) = 0, v (0, y) = y$ , giving $u_{y} (0, 0) = 0, v_{y} (0, 0) = 1$ . Check: $u_{x} = 1 = v_{y}$ ✅ and $u_{y} = 0 = - v_{x}$ ✅. CR holds at $0$ .

Step 2 — But the limit depends on direction. Take $Δ z = r e^{i θ}$ and let $r \to 0$ along a fixed ray. Since $f (0) = 0$ , $\frac{f ( Δ z ) - 0}{Δ z} = \frac{z ˉ ^{2} / z}{z} = \frac{z ˉ ^{2}}{z ^{2}} = \frac{( r e ^{- i θ} ) ^{2}}{( r e ^{i θ} ) ^{2}} = e^{- 4 i θ} .$ This value changes with the approach angle $θ$ : along $θ = 0$ it is $1$ ; along $θ = π /4$ it is $e^{- iπ} = - 1$ . Different directions give different answers, so the derivative limit does not exist. $f$ is not complex-differentiable at $0$ .

Step 3 — Which hypothesis broke? The sufficiency theorem needs CR plus continuity of the partials on a neighborhood. Here the partials of $u, v$ are not continuous at $0$ (the difference-quotient wobbles as $e^{- 4 i θ}$ ), so having CR only at the isolated point $0$ never earned us differentiability. This is the sharpest form of the parent's warning: CR at a point is necessary, not sufficient.