Complex Differentiability & the Cauchy–Riemann Equations

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Definition of Complex Differentiability

The concept of the derivative in complex analysis extends the familiar definition from real calculus, but the implications are far more profound due to the nature of the complex plane.

The derivative of a complex function f(z) at a point z₀ is defined by the limit f’(z₀) = $\lim_{h \to 0}\frac{f(z₀+h)-f(z₀)}{h}$, provided this limit exists and is independent of the path h takes to approach zero in the two-dimensional complex plane. This path independence is a much stronger condition than for real differentiability.

In real analysis, the variable $h$ can only approach zero from two directions: from the left or the right. In the complex plane, $h = h_1 + ih_2$ is a two-dimensional quantity, so it can approach zero along any curve — the real axis, the imaginary axis, spirals, zigzag paths, or any other trajectory. The requirement that a single, well-defined limit emerges from every such approach imposes a far more rigid constraint on $f$.

This definition can also be expressed as follows: Δz = z − z₀, f’(z₀) = $\lim_{Δz \to 0}\frac{f(z₀+Δz)-f(z₀)}{Δz}$. Typically, the subscript in z₀ is dropped for simplicity. Writing $w = f(z)$ and denoting the change in the function value corresponding to a change $\Delta z$ in the input by $\Delta w = f(z + \Delta z) - f(z)$, the derivative becomes: $\frac{dw}{dz} = \lim_{Δz \to 0}\frac{Δw}{Δz}$. The key aspect to consider is that the limit must exist and be the same regardless of the path h (or Δz) takes as it approaches zero in the complex plane.

The key aspect to bear in mind is that the limit must exist and be the same regardless of the path $h$ (or $\Delta z$) takes as it approaches zero in the complex plane.

If this limit exists, $f(z)$ is said to be complex differentiable at $z_0$, and the value of the limit is denoted by $f'(z_0)$. If a function is differentiable at every point in an open set $U \subseteq \mathbb{C}$, it is said to be analytic or holomorphic on $U$.

The Cauchy-Riemann Equations

The Cauchy-Riemann equations provide a necessary condition for a complex function $f(z) = u(x,y) + iv(x,y)$ to be differentiable at a point $z_0 = x_0 + iy_0$.

Suppose $f(z) = u(x,y) + iv(x,y)$ is differentiable at $z_0 = x_0 + iy_0$. By definition, the limit $f'(z_0) = \lim_{h \to 0}\frac{f(z_0+h)-f(z_0)}{h}$ must exist as a finite, unique complex number, irrespective of the path along which the complex increment $h$ approaches $0$.

We evaluate this limit along two canonical paths.

Path 1: Approach Along the Real Axis.

Set $h = h_1 + i \cdot 0$ where $h_1 \in \mathbb{R}$:

$$ \begin{aligned} f'(z_0) &= \lim_{h_1 \to 0} \frac{f((x_0+h_1)+iy_0)-f(x_0+iy_0)}{h_1} \\[2pt] &= \lim_{h_1 \to 0} \frac{\bigl[u(x_0+h_1, y_0)+iv(x_0+h_1, y_0)\bigr]-\bigl[u(x_0, y_0)+iv(x_0, y_0)\bigr]}{h_1} \\[2pt] &= \lim_{h_1 \to 0} \frac{u(x_0+h_1, y_0)-u(x_0, y_0)}{h_1}+i\lim_{h_1 \to 0} \frac{v(x_0+h_1, y_0) - v(x_0, y_0)}{h_1}\\[2pt] &= \frac{\partial u}{\partial x}(x_0, y_0) + i\frac{\partial v}{\partial x}(x_0, y_0). \end{aligned} $$

Path 2: Approach Along the Imaginary Axis

Set $h = 0 + i \cdot h_2$ where $h_2 \in \mathbb{R}$:

$$ \begin{aligned} f'(z_0) &= \lim_{h_2 \to 0} \frac{f(x_0+i(y_0+h_2))-f(x_0+iy_0)}{ih_2} \\[2pt] &= \lim_{h_2 \to 0} \frac{\bigl[u(x_0, y_0 + h_2)+iv(x_0, y_0 + h_2)\bigr] - \bigl[u(x_0, y_0) +iv(x_0, y_0)\bigr]}{ih_2}\\[2pt] &= \frac{1}{i}\lim_{h_2 \to 0} \frac{u(x_0, y_0 + h_2)- u(x_0, y_0)}{h_2} +\lim_{h_2 \to 0}\frac{v(x_0, y_0 + h_2)-v(x_0, y_0)}{h_2}\\[2pt] &= \frac{1}{i} \frac{\partial u}{\partial y}(x_0, y_0)+\frac{\partial v}{\partial y}(x_0, y_0) = -i\frac{\partial u}{\partial y}(x_0, y_0)+\frac{\partial v}{\partial y}(x_0, y_0). \end{aligned} $$

Since f(z) is differentiable at z₀, these two expressions for f(z₀) must be identical.

Equating real and imaginary parts: $\underbrace{\frac{\partial u}{\partial x}}_{\text{Re from Path 1}} + i\underbrace{\frac{\partial v}{\partial x}}_{\text{Im from Path 1}} = \underbrace{\frac{\partial v}{\partial y}}_{\text{Re from Path 2}} - i\underbrace{\frac{\partial u}{\partial y}}_{\text{Im from Path 2}}, which gives the Cauchy–Riemann equations: $\boxed{\frac{\partial u}{\partial x}(x_0, y_0) = \frac{\partial v}{\partial y}(x_0, y_0), \qquad \frac{\partial u}{\partial y}(x_0, y_0) = -\frac{\partial v}{\partial x}(x_0, y_0).}$

This derivation also provides two equivalent expressions for the complex derivative itself: (Real Axis) $f'(z) = \dfrac{\partial u}{\partial x} + i\dfrac{\partial v}{\partial x}$, (Imaginary Axis), f’(z) = $f'(z) = \dfrac{\partial v}{\partial y} - i\dfrac{\partial u}{\partial y}$

Relationship Between Complex Differentiability and the Cauchy-Riemann Equations

The relationship between complex differentiability and the Cauchy–Riemann equations is subtle: the equations are necessary but not sufficient on their own; an additional regularity condition (continuity of partial derivatives) is needed to make them sufficient.

Theorem: (Ncessary condition) If a complex function $f(z) = u(x, y) + iv(x, y)$ is complex-differentiable at a point $z_0 = x_0 + iy_0$, then its real and imaginary parts $u(x,y)$ and $v(x,y)$ exist and must satisfy the Cauchy–Riemann equations at that point.

This follows directly from the limit definition of the complex derivative. If f’(z₀) exists, then the limit f’(z₀) = $\lim_{h \to 0}\frac{f(z₀+h)-f(z₀)}{h}$ must exist and be the same regardless of the path h takes as it approaches zero. By considering two specific path: one along the real axis ($h = h_1$, $h_1 \in \mathbb{R}$) and another along the imaginary axis ($h = ih_2$, $h_2 \in \mathbb{R}$) and equating the resulting expression, the Cauchy-Riemann equations emerge.

If a function is genuinely complex-differentiable, the derivative limit must be consistent regardless of the path. Therefore, checking this consistency along these two canonical paths must yield the Cauchy-Riemann equations. This means the Cauchy-Riemann equations are a necessary condition for complex differentiability, but they do not, but themseles, guarantee consistency along all other possible paths in the complex plane. They provide a snapshot of the function’s behavior in two cardinal directions, but offer no inherent guarantee of the “smoothness” or “well-behavedness” (such as real differentiability or continuity of partial derivatives) required in a neighborhood of the point to ensure the path independence of the derivative.

Counterexample: A Function That Fails the Cauchy–Riemann Equations

• Let $f(z) = \bar{z} = x - iy$. The following example shows that a function can be extremely well-behaved from a real-variable standpoint and still fail to be complex differentiable, precisely because it violates the Cauchy–Riemann equations.

1. Continuity and Existence of Partial Derivatives. The function $f(z) = \bar{z}$ is continuous everywhere in $\mathbb{C}$. Its real and imaginary parts $u(x, y) = x$, $v(x, y) = -y$ have well-defined, continuous partial derivatives at every point: $\frac{\partial u}{\partial x} = 1, \frac{\partial u}{\partial y} = 0, \frac{\partial v}{\partial x} = 0, \frac{\partial v}{\partial y} = -1$
2. Failure of the Cauchy-Riemann Equations.Ckearkt, $\frac{\partial u}{\partial x}(x_0, y_0) = 1 \ne -1 = \frac{\partial v}{\partial y}(x_0, y_0)$. Therefore, the Cauchy-Riemann equations fails at every point.
3. Non-Differentiability: Since the Cauchy-Riemann equations fail everywhere, $f(z) = \bar{z}$ is nowhere complex-differentiable, despite its continuity and well-behaved partial derivatives.

One can also verify this directly from the definition. The difference quotient is: $\frac{f(z_0 + h) - f(z_0)}{h} = \frac{\bar{h}}{h}.$

• Along the real axis ($h = h_1 \in \mathbb{R}$): $\frac{\bar{h}}{h} = \frac{h_1}{h_1} = 1$.
• Along the imaginary axis ($h = ih_2$, $h_2 \in \mathbb{R}$): $\frac{\bar{h}}{h} = \frac{-ih_2}{ih_2} = -1$.

The limit depends on the direction, confirming non-differentiability.

Conclusion: Continuity alone, or even the existence and continuity of partial derivatives of u and v, is not sufficient for complex differentiability; the Cauchy-Riemann equations must hold.

This highlights two key points:

1. Necessity of Cauchy-Riemann: Complex differentiability requires more than just real-variable smoothness; it imposes specific structural constraints (the Cauchy-Riemann equations) linking u and v.
2. Path-Dependence of Complex Limits: The failure occurs because the complex derivative limit $\lim_{h \to 0}\frac{f(z₀+h)-f(z₀)}{h}$ depends on the direction of h (e.g., approaching along real or imaginary axes yields different results).

Sufficient Condition: Cauchy-Riemann Equations (with Continuous Partial Derivatives) Imply Complex Differentiability

While the Cauchy-Riemann equations are necessary for complex differentiability, they are not, on their own, sufficient to guarantee it.

Counterexamples

• Let $f(z) = \begin{cases} \frac{(\bar z)^2}{z}, &z \ne 0 \\\\ 0, &z = 0 \end{cases}$

Let z = x + iy, then its conjugate is $\bar{z} = x - iy$. So: $f(z) = \begin{cases} \frac{(x - iy)^2}{x + iy} &z \ne 0 \\\\ 0, &z = 0 \end{cases}$

Expanding: $f(z) = \begin{cases} \frac{x^2 - 2ixy - y^2}{x + iy} &x + iy \ne 0 \\\\ 0, &x + iy = 0 \end{cases} = \begin{cases} \frac{(x^2 - y^2) - 2ixy}{x + iy} &x + iy \ne 0 \\\\ 0, &x + iy = 0 \end{cases}$

Multiply numerator and denominator by the conjugate of the denominator: $f(z) = \begin{cases} \frac{(x^2 - y^2 - 2ixy)(x - iy)}{x^2 + y^2} &x + iy \ne 0 \\\\ 0, &x + iy = 0 \end{cases}$

Expanding and collecting real and imaginary parts:

$f(x + iy)= \begin{cases} \frac{x^3-3xy^2}{x^2+y^2} + i\frac{y^3-3x^2y}{x^2+y^2}, &z \ne 0 \\\\ 0, &z = 0 \end{cases}$

1. u = $\mathbb{Re(f)} = \frac{x^3-3xy^2}{x^2+y^2}$ (if z ≠ 0), u = 0 (if z = 0), v = $\mathbb{Im(f)} = \frac{y^3-3x^2y}{x^2+y^2}$ (if z ≠ 0), v = 0 (if z = 0).
2. Partial Derivatives at the Origin. $\frac{\partial u}{\partial x}(0, 0) = \lim_{h \to 0} \frac{u(h, 0) - u(0, 0)}{h} = \lim_{h \to 0} \frac{\frac{h^3}{h^2} - 0}{h} = 1$
   Likewise, $\frac{\partial u}{\partial y}(0, 0) = \lim_{h \to 0} \frac{u(0, h) - u(0, 0)}{h} = \lim_{h \to 0} \frac{0 - 0}{h} = 0$
   It is left to the reader to calculate, $\frac{\partial v}{\partial x}(0, 0) = 0, \frac{\partial v}{\partial y}(0, 0) = 1$
3. Check Cauchy-Riemann Equations: $\frac{\partial u}{\partial x}(0, 0) = 1 = \frac{\partial v}{\partial y}(0, 0) \text{ and } \frac{\partial u}{\partial y}(0, 0) = 0 = \frac{\partial v}{\partial x}(0, 0)$ at any point. Both equations are satisfied.
4. However, f is not differentiable at z = 0.

Path along the real axis ($h \in \mathbb{R}$): $\lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^3/h^2}{h} = 1.$

Path along the diagonal $h = t + it$, $t \to 0$, $t \in \mathbb{R}$: $\lim_{t \to 0} \frac{f(t+it)}{t+it} = \lim_{t \to 0} \frac{\frac{t^3-3t^3}{2t^2} + i\frac{t^3-3t^3}{2t^2}}{t+it} = \lim_{t \to 0} \frac{-t - it}{t+it} = -1.$

The two paths give different limits ($1$ versus $-1$), so $f$ is not complex-differentiable at $z = 0$, despite the Cauchy–Riemann equations being satisfied there. The Cauchy-Riemman equations are not sufficient to ensure complex differentiability at a point.

• Counterexample 2: $f(z) = e^{-1/z^4}$

Let $f(z) = \begin{cases} e^{-1/z^4}, & z \neq 0, \\ 0, & z = 0. \end{cases}$

1. Partial Derivatives at the Origin.
   Along the real axis ($z = h \in \mathbb{R}$): $f(h) = e^{-1/h^4}$, so $u(h, 0) = e^{-1/h^4}$ and $v(h, 0) = 0$.
   $\frac{\partial u}{\partial x}(0, 0) = \lim_{h \to 0} \frac{e^{-1/h^4}}{h} = 0 \quad \text{(exponential decay dominates)},$
   $\frac{\partial v}{\partial x}(0, 0) = \lim_{h \to 0} \frac{0}{h} = 0.$
   Along the imaginary axis ($z = ih$, $h \in \mathbb{R}$): $f(ih) = e^{-1/(ih)^4} = e^{-1/h^4}$, so $u(0, h) = e^{-1/h^4}$ and $v(0, h) = 0$.
   $\frac{\partial u}{\partial y}(0, 0) = \lim_{h \to 0} \frac{e^{-1/h^4}}{h} = 0, \qquad \frac{\partial v}{\partial y}(0, 0) = \lim_{h \to 0} \frac{0}{h} = 0.$
2. The Cauchy–Riemann Equations Hold at the Origin. $\frac{\partial u}{\partial x}(0, 0) = 0 = \frac{\partial v}{\partial y}(0, 0), \qquad \frac{\partial u}{\partial y}(0, 0) = 0 = -\frac{\partial v}{\partial x}(0, 0).$
3. However, $f$ Is Not Differentiable at $z = 0$. Let $h = re^{i\pi/4}$ where $r \in \mathbb{R}^+$ and $r \to 0$: $\frac{f(h) - f(0)}{h} = \frac{e^{-1/(re^{i\pi/4})^4}}{re^{i\pi/4}} = \frac{e^{-1/(r^4 e^{i\pi})}}{re^{i\pi/4}} = \frac{e^{+1/r^4}}{re^{i\pi/4}}.$

As $r \to 0^+$, $\frac{1}{e^{i\pi/4}}$ is just a constant, and $e^{1/r^4}/r \to \infty$. Thus, (i) along the real and imaginary axis, the limit is zero. (ii) along $h = re^{i\pi/4}$: the limit is $\infty$.

The limit does not exist, and $f$ is not complex-differentiable at $z = 0$.

Again, what went wrong? The partial derivatives exist at the origin and satisfy the Cauchy–Riemann equations, but they are not continuous there. The function $e^{-1/z^4}$ exhibits wild oscillatory and explosive behavior near the origin along certain directions in the complex plane, behavior that the partial derivatives along the two coordinate axes cannot detect.

Proposition. Let $f(z) = u(x, y) + iv(x, y)$ for $z = x + iy \in D$, where $D \subseteq \mathbb{C}$ is an open set. Assume that u and v have continuous first partial derivatives throughout D and they satisfy the Cauchy-Riemann equations at a point $z \in D$. Then, f'(z) exists, i.e., f is differentiable at z.

Proof.

Let $z = x + iy \in D$. Since $D$ is open, there exists $r > 0$ such that the open ball $B(z, r) = \{w \in \mathbb{C} : |w - z| < r\} \subseteq D$.

Let $h = p + iq$ be a complex increment with $|h| < r$, so that $z + h \in B(z, r) \subseteq D$.

1. Write the difference quotient. The difference quotient for the derivative of $f$ at $z$ is: $\frac{f(z+h)-f(z)}{h} = \frac{1}{h}\bigl[u(x+p,\, y+q)-u(x, y)+i\bigl(v(x+p,\, y+q)-v(x, y)\bigr)\bigr].$
   The goal is to show that as $h \to 0$, this quotient approaches a finite limit, proving $f'(z)$ exists.
2. Decompose the increments. To apply the Mean Value Theorem, split the differences in $u$ and $v$ by adding and subtracting intermediate terms (first varying $x$, then $y$):
   $\frac{f(z+h)-f(z)}{h} = \frac{1}{h}\bigl[\underbrace{u(x+p,\, y+q)-u(x,\, y + q)}_{\text{change in }x} + \underbrace{u(x,\, y + q) -u(x, y)}_{\text{change in }y}$
   $+ \; i\bigl(\underbrace{v(x+p,\, y+q)-v(x,\, y+q)}_{\text{change in }x} + \underbrace{v(x,\, y + q) - v(x, y)}_{\text{change in }y}\bigr)\bigr].$
   $= \frac{p}{h}\left(\frac{u(x+p,\,y+q)-u(x,\, y + q)}{p}+i\frac{v(x+p,\, y+q)-v(x,\,y+q)}{p}\right)$
   $+ \;\frac{q}{h}\left(\frac{u(x,\, y + q) -u(x, y)}{q}+i\frac{v(x,\, y + q) - v(x, y)}{q}\right).$
3. Apply the Mean Value Theorem. Since $u$ and $v$ have continuous first partial derivatives, the Mean Value Theorem for real functions applies to each difference quotient.
   Recall: Mean Value Theorem for Functions of Several Variables. Let $g: \mathbb{R}^2 \to \mathbb{R}$ be differentiable on an open set containing the line segment joining $\mathbf{a}$ and $\mathbf{b}$. Then, there exists a point $\mathbf{c}$ on that segment such that $g(\mathbf{b}) - g(\mathbf{a}) = \nabla g(\mathbf{c}) \cdot (\mathbf{b} - \mathbf{a}).$
   The average rate of change between two points is achieved somewhere in between.
   For the $x$-direction terms (with $y + q$ fixed), there exist $\alpha_1, \alpha_2 \in (0, 1)$ such that: $\frac{u(x+p,\,y+q)-u(x,\, y + q)}{p} = \frac{\partial u}{\partial x}(x+ \alpha_1 p,\, y + q),$
   $\frac{v(x+p,\, y+q)-v(x,\,y+q)}{p} = \frac{\partial v}{\partial x}(x+ \alpha_2 p,\, y + q).$
   For the $y$-direction terms (with $x$ fixed), there exist $\alpha_3, \alpha_4 \in (0, 1)$ such that: $\frac{u(x,\, y + q) -u(x, y)}{q} = \frac{\partial u}{\partial y}(x,\, y+ \alpha_3 q),$
   $\frac{v(x,\, y + q) - v(x, y)}{q} = \frac{\partial v}{\partial y}(x,\, y+ \alpha_4 q).$
4. Introduce error terms using continuity. Since the partial derivatives are continuous at $(x, y)$, as $h \to 0$ (so $p \to 0$ and $q \to 0$), the intermediate points $(x + \alpha_1 p,\, y + q)$, $(x + \alpha_2 p,\, y + q)$, $(x,\, y + \alpha_3 q)$, and $(x,\, y + \alpha_4 q)$ all converge to $(x, y)$.
   Thus, each partial derivative can be written as its value at $(x, y)$ plus a vanishing error: $\frac{\partial u}{\partial x}(x+ \alpha_1 p, y + q) = \frac{\partial u}{\partial x}(x, y) + \epsilon_1, \qquad \frac{\partial v}{\partial x}(x+ \alpha_2 p, y + q) = \frac{\partial v}{\partial x}(x, y) + \epsilon_2,$
   $\frac{\partial u}{\partial y}(x, y+ \alpha_3 q) = \frac{\partial u}{\partial y}(x, y) + \epsilon_3, \qquad \frac{\partial v}{\partial y}(x, y+ \alpha_4 q) = \frac{\partial v}{\partial y}(x, y) + \epsilon_4,$
   where $\epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4 \to 0$ as $h = p + iq \to 0$.
5. Substitute and group. Substituting into the difference quotient: $\frac{f(z+h)-f(z)}{h} = \frac{p}{h}\!\left(\frac{\partial u}{\partial x}(x, y) + \epsilon_1 + i\!\left(\frac{\partial v}{\partial x}(x, y) + \epsilon_2\right)\!\right)$
   $+ \;\frac{q}{h}\!\left(\frac{\partial u}{\partial y}(x, y)+ \epsilon_3 +i\!\left(\frac{\partial v}{\partial y}(x, y) + \epsilon_4\right)\!\right).$
6. Apply the Cauchy–Riemann equations. Using $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$, substitute into the $y$-direction term:
   $\frac{f(z+h)-f(z)}{h} = \frac{p}{h}\!\left(\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}\right) + \frac{q}{h}\!\left(-\frac{\partial v}{\partial x} + i\frac{\partial u}{\partial x}\right) + \frac{p}{h}(\epsilon_1 + i\epsilon_2) + \frac{q}{h}(\epsilon_3 + i\epsilon_4),$
   where all partial derivatives are evaluated at $(x, y)$. Factoring the main terms:
   $= \frac{\partial u}{\partial x}\!\left(\frac{p}{h} + i\frac{q}{h}\right) + i\frac{\partial v}{\partial x}\!\left(\frac{p}{h} + i\frac{q}{h}\right) + \frac{p}{h}(\epsilon_1 + i\epsilon_2) + \frac{q}{h}(\epsilon_3 + i\epsilon_4).$
7. Simplify. Observe the key simplification: $\frac{p}{h} + i\frac{q}{h} = \frac{p + iq}{h} = \frac{h}{h} = 1.$
   Therefore, $\frac{f(z+h)-f(z)}{h} = \frac{\partial u}{\partial x}(x, y) + i\frac{\partial v}{\partial x}(x, y) + \underbrace{\frac{p}{h}(\epsilon_1 + i\epsilon_2) + \frac{q}{h}(\epsilon_3 + i\epsilon_4)}_{\text{error term } E(h)}.$
   The first term $\frac{\partial u}{\partial x}(x, y) + i\frac{\partial v}{\partial x}(x, y)$ is independent of $h$.
8. We ain to show that the error term vanishes. It remains to prove that $E(h) \to 0$ as $h \to 0$. Consider a typical term, say $\frac{p}{h}\epsilon_1$:
   $\left|\frac{p}{h}\epsilon_1\right| = \frac{|p|}{|h|} \cdot |\epsilon_1|.$
   Since $h = p + iq$, we have $|h| = \sqrt{p^2 + q^2} \geq |p|$, and therefore: $\frac{|p|}{|h|} \leq 1.$
   As $h \to 0$, continuity of $\frac{\partial u}{\partial x}$ guarantees $\epsilon_1 \to 0$, so:
   $\left|\frac{p}{h}\epsilon_1\right| \leq |\epsilon_1| \to 0.$
   By the same argument (using $|q|/|h| \leq 1$), each of the four error terms vanishes:
   $\frac{p}{h}\epsilon_1 \to 0, \quad \frac{p}{h}\epsilon_2 \to 0, \quad \frac{q}{h}\epsilon_3 \to 0, \quad \frac{q}{h}\epsilon_4 \to 0.$
   Hence $E(h) \to 0$.
9. Conclusion. Taking the limit: $\lim_{h \to 0}\frac{f(z+h)-f(z)}{h} = \frac{\partial u}{\partial x}(x, y) + i\frac{\partial v}{\partial x}(x, y).$

The limit exists, is finite, and is independent of the path $h$ takes to zero (since no assumption about the path was used — the bound $|p|/|h| \leq 1$ holds universally). Therefore, $f$ is differentiable at $z$, and

$$\boxed{f'(z) = \frac{\partial u}{\partial x}(x, y) + i\frac{\partial v}{\partial x}(x, y).} \qquad \blacksquare$$

The continuity condition bridges the gap between a pointwise algebraic relation (the Cauchy–Riemann equations checked along two directions) and the full path-independent limit required for complex differentiability. It ensures that the linear approximation provided by the partial derivatives is uniform across all directions, which is precisely what makes the error term vanish regardless of the path that takes h to zero.