advanced27 min studyLesson 361 in Mathematics

Complex differentiability and the Cauchy–Riemann equations

Recognize a complex derivative as one real-linear approximation compatible with every approach direction.

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What you’ll understand

  • Distinguish complex differentiability at a point from holomorphicity on an open set.
  • Derive the Cauchy–Riemann equations from directional difference quotients.
  • Use real differentiability and the Cauchy–Riemann equations with their correct hypotheses.

A function of one complex variable is also a function of two real coordinates, but a complex derivative asks for much more than two partial derivatives. One complex number must describe the first-order change for every direction of a small complex displacement. That requirement excludes most smooth planar maps and ultimately leads to the rigidity of holomorphic functions. Begin by distinguishing a local derivative at one point from holomorphicity throughout an open neighborhood; confusing them makes a pointwise equation test appear stronger than it is.

One multiplier must work in every direction

Let f be defined on an open neighborhood of a complex point z₀. Its derivative is the limit of (f(z0+h)−f(z0))/h(f(z_0+h)-f(z_0))/h as nonzero complex h tends to zero through arbitrary directions. The same limit must hold for horizontal, vertical, and curved approaches. For f(z)=z2f(z)=z^2, expansion gives the quotient 2z0+h2z_0+h, which tends to 2z02z_0 independently of direction. For conjugation, the quotient is h‾/h\overline h/h, which equals one on real increments and minus one on purely imaginary increments.

Multiplication by a complex number α is a very special real-linear transformation. If α=a+ib, it scales all lengths by its modulus and, when α is nonzero, rotates every direction by its argument. Its real matrix has equal diagonal entries and opposite off-diagonal entries. A general Jacobian can shear or stretch two directions differently and therefore cannot represent complex multiplication. The zero multiplier is allowed for differentiability, although it does not preserve angles as a nondegenerate local map.

f′(z0)=lim⁡h→0f(z0+h)−f(z0)h,h↦(a+ib)h ⟷ (a−bba)f'(z_0)=\lim_{h\to0}\frac{f(z_0+h)-f(z_0)}h,\qquad h\mapsto(a+ib)h\ \longleftrightarrow\ \begin{pmatrix}a&-b\\b&a\end{pmatrix}
The limit uses arbitrary complex increments. A complex derivative is a complex-linear first-order approximation, not an arbitrary two-by-two real Jacobian.↔ Scroll wide equations horizontally.

Use the Cauchy–Riemann equations with regularity

Write f(x+iy)=u(x,y)+iv(x,y)f(x+iy)=u(x,y)+iv(x,y). Horizontal increments identify the derivative with ux+ivxu_x+iv_x, while vertical increments identify it with vy−iuyv_y-iu_y. Equality yields the Cauchy–Riemann equations. Complex differentiability implies real differentiability and these equations at the point. In the converse direction, real differentiability at that point together with the equations is sufficient: the real-linear approximation then has exactly the complex-multiplication form, and its remainder is small compared with the increment length.

A convenient stronger hypothesis is that u and v have continuous first partial derivatives on an open set. If the equations hold throughout that set, the map is complex differentiable there and hence holomorphic. Existence of partial derivatives and a pointwise algebraic check alone is not that theorem; partial derivatives need not assemble into a valid first-order approximation. Check regularity and the set on which the equations hold before drawing a conclusion about an open neighborhood.

ux=vy,uy=−vx,f′(z)=ux+ivxwhen the complex derivative existsu_x=v_y,\qquad u_y=-v_x,\qquad f'(z)=u_x+iv_x\quad\text{when the complex derivative exists}
Real differentiability plus the two equations at a point is sufficient there. Continuous first partials and the equations throughout an open set give holomorphicity on that set.↔ Scroll wide equations horizontally.

An isolated derivative is not an analytic region

The function f(z)=∣z∣2f(z)=|z|^2 has derivative zero at the origin because its difference quotient there is h‾\overline h, which tends to zero. Away from the origin its Cauchy–Riemann equations fail. Thus the function is differentiable at one point but holomorphic on no nonempty open set. This example separates the definition of a derivative from the neighborhood condition needed for Cauchy theory, power-series representations, and the other global consequences developed later.

Holomorphic means complex differentiable at every point of an open set. A theorem of complex analysis then shows that the function has derivatives of every order and a convergent local power series; those conclusions are not extra assumptions hidden in the definition. For the moment, a reliable workflow is to state the open domain, inspect singularities or conjugates, establish real regularity, and test the equations there. This preserves both the power and the limits of the coordinate calculation.

Check a polynomial by both definitions

For f(z)=z2f(z)=z^2, calculate the derivative at z0=1+iz_0=1+i directly and through its real Jacobian. Identify the associated local rotation and scale.

  1. Expansion gives (f(z0+h)−f(z0))/h=2z0+h(f(z_0+h)-f(z_0))/h=2z_0+h, so the complex derivative is 2+2i2+2i. This calculation already handles all nonzero approach directions.

  2. The real and imaginary parts are u=x2−y2u=x^2-y^2 and v=2xyv=2xy. At (1,1), their Jacobian is (2−222)\begin{pmatrix}2&-2\\2&2\end{pmatrix}, satisfying both Cauchy–Riemann equations.

  3. The multiplier 2+2i2+2i has modulus 222\sqrt2 and argument π/4\pi/4. Thus the first-order map scales by 222\sqrt2 and rotates by 45 degrees; the quadratic remainder is of smaller order than the increment.

The result

Both approaches give f′(1+i)=2+2if'(1+i)=2+2i. The Jacobian's special form, rather than its mere existence, makes this a complex derivative.

Explore the model

Interactive lab

Interactive lab · matrix

Test a real Jacobian against complex linearity

Pause and Reset; set Diagonal scale a=1 and Shear b=0 to obtain the identity. Predict that it has the complex-multiplication form. For a fresh comparison Reset and set a=1,b=1: the off-diagonal entries now violate the Cauchy–Riemann pattern. Inspect the sheared grid. This family illustrates real linear maps; it does not evaluate an arbitrary complex function or test its neighborhood regularity.

Test a real Jacobian against complex linearityThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA smooth real map becomes complex differentiable only when its first-order transformation has the required special form.

Check your understanding

What is sufficient for complex differentiability at one point?

Try it yourself

Determine every point where f(z)=∣z∣2f(z)=|z|^2 is complex differentiable and whether it is holomorphic on any nonempty open set.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use u=x²+y² and v=0 to find possible points, then verify the difference quotient at the remaining candidate.

Compare with the solution

The equations require 2x=02x=0 and 2y=02y=0, so only the origin is possible. There, (∣h∣2−0)/h=h‾→0(|h|^2-0)/h=\overline h\to0, proving derivative zero. No nonempty open set consists only of the origin, so the function is holomorphic on none. This shows why one successful pointwise derivative does not imply a local power-series theory.

Take it with you

  • A complex derivative is one multiplier for all increment directions.
  • The Cauchy–Riemann converse needs a real differentiability or suitable regularity hypothesis.
  • Holomorphicity is an open-set condition, stronger than differentiability at an isolated point.

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