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Now is the time for us to shine. The time when our dreams are within reach and possibilities vast. Now is the time for all of us to become the people we have always dreamed of being. This is your world. You’re here. You matter. The world is waiting, Haley James Scott.
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.
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 and the change in the function value w = f(z) corresponding to a change Δz in the input is denoted by Δw = f(z + Δz) − f(z). Hence, the derivative could be written as $\frac{dw}{dz} = \lim_{Δz \to 0}\frac{Δw}{Δz}$. 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.
If this limit exists, f(z) it is said to be complex differentiable at z₀ and the value of the limit is denoted by f’(z₀). If a function is differentiable at every point in an open set U ⊆ ℂ it is said to be analytic or holomorphic on U.
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 z₀ = x₀ + iy₀.
Suppose f(z) = u(x,y) + iv(x,y) is differentiable at z₀ = x₀ + iy₀. By definition, the limit f’(z₀) = $\lim_{h \to 0}\frac{f(z₀+h)-f(z₀)}{h}$ must exist as a finite, unique complex number, irrespective of the path along which the complex increment h approaches 0.
Path 1: Approach along the real axis, h = h₁ + i·0 where h₁ ∈ ℝ, f’(z₀) = $\lim_{h \to 0}\frac{f(z₀+h)-f(z₀)}{h} = \lim_{h_1 \to 0} \frac{f((x_0+h_1)+iy_0)-f(x_0+iy_0)}{h_1} = \lim_{h_1 \to 0} \frac{u(x_0+h_1, y_0)+iv(x_0+h_1, y_0)-u(x_0, y_0)-iv(x_0, y_0)}{h_1} = \lim_{h_1 \to 0} \frac{u(x_0+h_1, y_0)-u(x_0, y_0)}{h_1}+i\lim_{h_1 \to 0} \frac{v(x_0+h_1, y_0) - v(x_0, y_0)}{h_1} = \frac{\partial u}{\partial x}(x_0, y_0) + i\frac{\partial v}{\partial x}(x_0, y_0)$
Path 2: Approach along the imaginary axis, h = 0 + i·h₂ where h₂∈ ℝ, f’(z₀) = $\lim_{h \to 0}\frac{f(z₀+h)-f(z₀)}{h} = \lim_{h_2 \to 0} \frac{f(x_0+i(y_0+h_2))-f(x_0+iy_0)}{ih_2} = \lim_{h_2 \to 0} \frac{u(x_0, y_0 + h_2)+iv(x_0, y_0 + h_2) - u(x_0, y_0) -iv(x_0, y_0)}{ih_2} = \lim_{h_2 \to 0} \frac{u(x_0, y_0 + h_2)- u(x_0, y_0)}{ih_2} +i\lim_{h_2 \to 0}\frac{v(x_0, y_0 + h_2)-v(x_0, y_0)}{ih_2} = \frac{1}{i} \frac{\partial u}{\partial y}(x_0, y_0)+\frac{\partial v}{\partial y}(x_0, y_0) = -i\frac{\partial u}{\partial y}(x_0, y_0)+\frac{\partial v}{\partial y}(x_0, y_0)$
Since f(z) is differentiable at z₀, these two expressions for f(z₀) match exactly (real and imaginary parts separately), which gives the Cauchy-Riemann equations, $\frac{\partial u}{\partial x}(x_0, y_0) = \frac{\partial v}{\partial y}(x_0, y_0), \frac{\partial u}{\partial y}(x_0, y_0) = -\frac{\partial v}{\partial x}(x_0, y_0)$.
This derivation also provides expressions for the complex derivative itself: (Real Axis) f’(z) = $\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}$, (Imaginary Axis), f’(z) = $\frac{\partial v}{\partial y} - i\frac{\partial u}{\partial y}$
If a complex function f(z) = u(x, y) + iv(x, y) is complex-differentiable at a point z₀ =x₀ +iy₀, then its real and imaginary parts, u(x,y) and v(x,y), must satisfy the Cauchy-Riemann equations at that point. This is a necessary condition for complex differentiability. 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 = h₁ + i·0 where h₁ ∈ ℝ and another along the imaginary axis, h = 0 + i·h₂ where h₂∈ 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 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.
If a function f(z) = u(x, y) + iv(x, y) is complex-differentiable at a point z₀ =x₀ +iy₀, then its real and imaginary parts, u(x,y) and v(x,y), must satisfy the Cauchy-Riemann equations at that point: $\frac{\partial u}{\partial x}(x_0, y_0) = \frac{\partial v}{\partial y}(x_0, y_0), \frac{\partial u}{\partial y}(x_0, y_0) = -\frac{\partial v}{\partial x}(x_0, y_0)$.
1️⃣ Continuity and Existence of Partial Derivatives: $f(z) = \bar{z}$ is continuous everywhere in ℂ. Its real and imaginary parts u(x, y) = x, v(x, y) = -y have well-defined, continuous partial derivatives at every point: $\frac{\partial u}{\partial x} = 1, \frac{\partial u}{\partial y} = 0, \frac{\partial v}{\partial x} = 0, \frac{\partial v}{\partial y} = -1$
2️⃣ Failure of the Cauchy-Riemann Equations.Obviously, $\frac{\partial u}{\partial x}(x_0, y_0) = 1 \ne \frac{\partial v}{\partial y}(x_0, y_0) = -1, \frac{\partial u}{\partial y}(x_0, y_0) = 0 = -\frac{\partial v}{\partial x}(x_0, y_0)$. Therefore, the Cauchy-Riemann equations are not satisfied at any point.
3️⃣ 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.
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:
While the Cauchy-Riemann equations are necessary for complex differentiability, they are not, on their own, sufficient to guarantee it.
Let z = x + iy, then its conjugate is $\bar{z} = x - iy$. So: $f(z) = \begin{cases} \frac{(x - iy)^2}{x + iy} &x + i \ne 0 \\\\ 0, &z = 0 \end{cases}$
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: $f(z) = \begin{cases} \frac{(x^2 - y^2 - 2ixy)(x - iy)}{x^2 + y^2} &x + iy \ne 0 \\\\ 0, &x + iy = 0 \end{cases}$
$f(x + iy)= \begin{cases} \frac{x^3-3xy^2}{x^2+y^2} + i\frac{y^3-3x^2y}{x^2+y^2}, &z \ne 0 \\\\ 0, &z = 0 \end{cases}$
1️⃣ Continuity and Existence of Partial Derivatives:
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).
$\frac{\partial u}{\partial x}(0, 0) = \lim_{h \to 0} \frac{u(h, 0) - u(0, 0)}{h} = \lim_{h \to 0} \frac{\frac{h^3}{h^2} - 0}{h} = 1$
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, $\frac{\partial v}{\partial x}(0, 0) = 0, \frac{\partial v}{\partial y}(0, 0) = 1$
2️⃣ Check Cauchy-Riemann Equations: $\frac{\partial u}{\partial x}(0, 0) = 1 = \frac{\partial v}{\partial y} \text{ and } \frac{\partial u}{\partial y}(0, 0) = 0 = \frac{\partial v}{\partial x}$ at any point.
3️⃣ However, f is not differentiable at z = 0.
If h ∈ ℂ approaches zero through real numbers then, $\lim_{h \to 0, h \in ℝ} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{\frac{h^3}{h^2} - 0}{h} = 1$
If h ∈ ℂ approaches zero through numbers of the form t + it where t → 0.
$\lim_{t \to 0, t \in ℝ} \frac{f(0+t+it) - f(0)}{t+it} = \lim_{t \to 0, t \in ℝ} \frac{\frac{t^3-3t^3}{2t^2} + i\frac{t^3-3t^3}{2t^2}}{t+it} = \lim_{t \to 0, t \in ℝ} \frac{-t-it}{t+it} = \lim_{t \to 0, t \in ℝ} -1 = -1$
Therefore, we have two paths with different directions approaching zero that give different limits, therefore f is not differentiable at zero. The Cauchy-Riemman equations are not sufficient to ensure complex differentiability at a point.
Partial derivatives: Along the real axis (h ∈ ℝ), f(h) = exp(-1/h⁴), u(h, 0) = exp(-1/h⁴), v(h, 0) = 0. $\frac{\partial u}{\partial x}(0, 0) = \lim_{h \to 0} \frac{u(h, 0) -u(0, 0)}{h} = \lim_{h \to 0} \frac{e^{-1/h^4}}{h}$ =0 (exponential decay dominates h), $\frac{\partial v}{\partial x}(0, 0) = \lim_{h \to 0} \frac{v(h, 0) -u(0, 0)}{h} = \lim_{h \to 0} \frac{0}{h} = 0.$
Along the imaginary axis. f(ih) = exp(-1/(ih)⁴) = exp(-1/h⁴), u(0, h) = exp(-1/h⁴), v(0, h) = 0. $\frac{\partial u}{\partial y}(0, 0) = \lim_{h \to 0} \frac{u(0, h) -u(0, 0)}{ih} = \lim_{h \to 0} \frac{e^{-1/h^4}}{ih}$ =0 (exponential decay dominates h), $\frac{\partial v}{\partial y}(0, 0) = \lim_{h \to 0} \frac{v(0, h) -u(0, 0)}{ih} = \lim_{h \to 0} \frac{0}{ih} = 0.$
Result: $\frac{\partial u}{\partial x}(0, 0) = 0 = \frac{\partial v}{\partial y}(0, 0), \frac{\partial u}{\partial y}(0, 0) = 0 = -\frac{\partial v}{\partial x}(0, 0)$. The Cauchy-Riemann equations hold at z = 0. However, f is not complex-differentiable at z = 0. Let h = $re^{i\pi/4}$ where r ∈ ℝ+ and r → 0.
$f'(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h} =[\text{In particular, }] = \lim_{r \to 0^+} \frac{exp(-1/(re^{i\pi/4})^4)}{re^{i\pi/4}} = \lim_{r \to 0^+} \frac{exp(-1/r^4e^{\pi·i})}{re^{i\pi/4}} = \lim_{r \to 0^+} \frac{exp(1/r^4)}{r e^{i\pi/4}} = \frac{1}{e^{i\pi/4}} \lim_{r \to 0^+} \frac{exp(1/r^4)}{r} = ∞$. The limit depends on the path: it is 0 along the real and imaginary axis but ∞ along the diagonal. Thus, $\lim_{h \to 0} \frac{f(h)-f(0)}{h}$ does not exist, and f is not complex-differentiable at z = 0.
This example illustrates that the mere existence of partial derivatives satisfying the Cauchy-Riemann equations is insufficient. However, if we add the condition that the first-order partial derivatives of u and v not only exist in a neighborhood of z₀ but are also continuous at z₀ and satisfy the Cauchy-Riemann equations at that very point, then the function is guaranteed to be complex-differentiable at z₀. Thus, the combination of continuous partial derivatives and the Cauchy-Riemann Equations equations forms a sufficient condition for complex differentiability.
If the partial derivatives of u and v, i.e., $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}$ not only exist in a neighborhood of z₀ = x₀ + iy₀, but are also continuous in a neighborhood of z₀ and they satisfy the Cauchy-Riemann equations at z₀, then f(z) = u(x, y)+ iv(x, y) is complex-differentiable at z₀. This continuity condition is crucial because it ensures that the function behaves “nicely” in the vicinity of the point, allowing for a uniform linear approximation.
Proposition. Let f(z) = u(x, y) + iv(x, y) for z = x + iy ∈ D where D ⊆ 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 ∈ D. Then, f'(z) exists, i.e., f is differentiable at z.
Proof.
Let z = z = x + iy ∈ D. Since D is a open set, there exists r > 0 such that the open ball B(z, r) = {w ∈ ℂ: |w -z| < r}.
Let h = p + iq be a complex increment with |h| < r, so z + h ∈ B(z, r) ⊆ D.
The difference quotient for the derivative of f at z is:
$\frac{f(z+h)-f(z)}{h} = \frac{1}{h}[u(x+p,y+q)-u(x, y)+i(v(x+p, y+q)-v(x, y))]$. The goal is to show that as h→0, this quotient approaches a finite limit, proving f′(z) exists.
To apply the Mean Value Theorem, the differences in u and v are split by adding and subtracting intermediate terms:
$\frac{f(z+h)-f(z)}{h} = \frac{1}{h}[u(x+p,y+q)-u(x, y + q) + u(x, y + q) -u(x, y)+i(v(x+p, y+q)-v(x, y+q) + v(x, y + q) - v(x, y))] = \frac{p}{h}(\frac{u(x+p,y+q)-u(x, y + q)}{p}+i\frac{v(x+p, y+q)-v(x, y+q)}{p})+\frac{q}{h}(\frac{u(x, y + q) -u(x, y)}{q}+i\frac{v(x, y + q) - v(x, y)}{q})$
Since u and v have continuous first partial derivatives, the Mean Value Theorem for real functions of two variables is applied to each difference quotient: For the x-direction terms (fixed y+q): there exists α₁ ∈ (0, 1) such that: $\frac{u(x+p,y+q)-u(x, y + q)}{p}= \frac{\partial u}{\partial x}(x+ \alpha_1p, y + q)$. Likewise, there exists α₂ ∈ (0, 1) such that $\frac{v(x+p, y+q)-v(x, y+q)}{p} = \frac{\partial v}{\partial x}(x+ \alpha_2p, y + q)$
Recall. Let f: $\mathbb{R}^2 \to \mathbb{R}$ be a differentiable function. Suppose f is defined and differentiable on an open set containing the line segment joining two points $\mathbf{a} = (x_1, y_1)$ and $\mathbf{b} = (x_2, y_2)$. Then: $f(\mathbf{b}) - f(\mathbf{a}) = \nabla f(\mathbf{c}) \cdot (\mathbf{b} - \mathbf{a})$ for some point $\mathbf{c}$ on the line segment between a and b, where:
For the y-direction terms (fixed x): there exist α₃, α₄ such that, $\frac{u(x, y + q) -u(x, y)}{q} = \frac{\partial u}{\partial y}(x, y+ \alpha_3q), \frac{v(x, y + q) - v(x, y)}{q} = \frac{\partial v}{\partial y}(x, y+ \alpha_4q)$
Since the partial derivatives are continuous at (x, y), as h →0 (so p→0 and q→0), the points (x + α₁p, y+ q), (x + α₂p, y + q), etc., all approach (x, y). Thus, each partial derivative can be expressed as its value at (x, y) plus an error term that vanishes as h →0:
$\frac{\partial u}{\partial x}(x+ \alpha_1p, y + q) = \frac{\partial u}{\partial x}(x, y) + \epsilon_1$. Similarly, $\frac{\partial v}{\partial x}(x+ \alpha_2p, y + q) = \frac{\partial v}{\partial x}(x, y) + \epsilon_2, \frac{\partial u}{\partial y}(x, y+ \alpha_3q) = \frac{\partial u}{\partial y}(x, y)+ \epsilon_3, \frac{\partial v}{\partial y}(x, y+ \alpha_4q) = \frac{\partial v}{\partial y}(x, y) + \epsilon_4$ where $\epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4 \to 0 \text{ as } h = p + iq \to 0$
Substituting these into the expression:
$\frac{f(z+h)-f(z)}{h} = \frac{p}{h}(\frac{\partial u}{\partial x}(x, y) + \epsilon_1 + i(\frac{\partial v}{\partial x}(x, y) + \epsilon_2)) +\frac{q}{h}(\frac{\partial u}{\partial y}(x, y)+ \epsilon_3 +i(\frac{\partial v}{\partial y}(x, y) + \epsilon_4))$
Applying the Cauchy-Riemann Equations: $\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} \text{ and } \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$.
Substitute these into the y-direction term then:
$\frac{f(z+h)-f(z)}{h} = \frac{p}{h}(\frac{\partial u}{\partial x}(x, y) + i(\frac{\partial v}{\partial x}(x, y))) + \frac{q}{h}(\frac{-\partial v}{\partial x}(x, y) +i(\frac{\partial u}{\partial x}(x, y))) + \frac{p}{h}(\epsilon_1 + i\epsilon_2) + \frac{q}{h} (\epsilon_3 + i\epsilon_4) = \frac{\partial u}{\partial x}(x, y)(\frac{p}{h}+i\frac{q}{h})+i\frac{\partial v}{\partial x}(x, y)(\frac{p}{h}+i\frac{q}{h})+ \frac{p}{h}(\epsilon_1 + i\epsilon_2) + \frac{q}{h} (\epsilon_3 + i\epsilon_4)$
Simplify the coefficients: $\frac{p}{h}+i\frac{q}{h} = \frac{p + iq}{h} = \frac{h}{h} = 1$, $\frac{f(z+h)-f(z)}{h} = \frac{\partial u}{\partial x}(x, y)+i\frac{\partial v}{\partial x}(x, y) + \frac{p}{h}(\epsilon_1 + i\epsilon_2) + \frac{q}{h} (\epsilon_3 + i\epsilon_4)$
The term $\frac{\partial u}{\partial x}(x, y)+i\frac{\partial v}{\partial x}(x, y)$ is independent of h.
$\lim_{h \to 0}\frac{f(z+h)-f(z)}{h} = \frac{\partial u}{\partial x}(x, y)+i\frac{\partial v}{\partial x}(x, y) + \lim_{h \to 0}\frac{p}{h}(\epsilon_1 + i\epsilon_2) + \frac{q}{h} (\epsilon_3 + i\epsilon_4)$
$\lim_{h \to 0} |\frac{p}{h}\epsilon_1| = lim_{h \to 0} |\frac{p}{h}||\epsilon_1| \le lim_{h \to 0} |\epsilon_1|$ since $\frac{|p|}{|h|} \le 1$ because h = p + iq, $|h| = \sqrt{p^2+q^2} \ge |p|$, and as h → 0, because the partial derivative is continuos ε₁ → 0, and $lim_{h \to 0} |\frac{p}{h}||\epsilon_1| = 0$ . Therefore, $\lim_{h \to 0} |\frac{p}{h}\epsilon_1| = 0 \leadsto \lim_{h \to 0} \frac{p}{h}\epsilon_1 = 0$. Similarly with the other limits, we could reach the conclusion: $\lim_{h \to 0}\frac{f(z+h)-f(z)}{h} = \frac{\partial u}{\partial x}(x, y)+i\frac{\partial v}{\partial x}(x, y)$, and the complex combination $\frac{p}{h}(\epsilon_1 + i\epsilon_2) + \frac{q}{h} (\epsilon_3 + i\epsilon_4) \to 0.$
Therefore: $\lim_{h \to 0}\frac{f(z+h)-f(z)}{h} = \frac{\partial u}{\partial x}(x, y)+i\frac{\partial v}{\partial x}(x, y)$. The limit exists and equals $ \frac{\partial u}{\partial x}(x, y)+i\frac{\partial v}{\partial x}(x, y)$, so f is differentiable at z, and f′(z) = $\frac{\partial u}{\partial x}(x, y)+i\frac{\partial v}{\partial x}(x, y)$
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