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FALSE: the Cauchy–Riemann equations at one point imply complex differentiability there

Statement

False claim: if the four first coordinate partial derivatives of $f = u + i v$ exist at a point and satisfy $u_{x} = v_{y}$ and $u_{y} = - v_{x}$ there, then $f$ is complex differentiable at that point.

Facts & Assumptions

Given: The function $f (0) = 0, f (z) = \frac{{\overset{ˉ}{z}}^{2}}{z} (z \neq 0) .$

[L1]

Complex differentiability at $0$ requires a single limit of $(f (h) - f (0)) / h$ as nonzero complex $h$ tends to $0$ (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

[L2]

Complex differentiability is equivalent to real total differentiability plus the Cauchy–Riemann equations; the equations alone are not asserted to be sufficient (Complex differentiability is equivalent to real total differentiability together with a complex-linear derivative, with $\partial_{\overset{ˉ}{z}} f = 0$ , or with the Cauchy–Riemann equations).

[L3]

The modulus is multiplicative, $z \overline{z} = ∣ z ∣^{2}$ , $\overline{\overline{z}} = z$ , and $∣ z ∣ = 0$ exactly when $z = 0$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive). Since $\overline{\overline{z}} = z$ and $z \overline{z} = ∣ z ∣^{2}$ , one has $∣ \overline{z} ∣^{2} = \overline{z} \overline{\overline{z}} = \overline{z} z = ∣ z ∣^{2}$ , and both moduli are nonnegative, so $∣ \overline{z} ∣ = ∣ z ∣$ .

Refutation

technique · direct counterexample

1.1

For $z \neq 0$ , $∣ f (z) ∣ = ∣ \overset{ˉ}{z} ∣^{2} / ∣ z ∣ = ∣ z ∣$ . Hence $f (z) \to 0 = f (0)$ as $z \to 0$ , so the example is even continuous at the point in question.

L3algebra

1.2

On the real axis, $f (t) = t$ ; on the imaginary axis, $f (i t) = i t$ . Therefore at the origin $u_{x} = 1, v_{x} = 0, u_{y} = 0, v_{y} = 1.$

givenalgebra

1.3

For $z \neq 0$ , the derivative quotient at the origin is $\frac{f (z) - f (0)}{z} = \frac{{\overset{ˉ}{z}}^{2}}{z^{2}} .$

givenalgebra

2.1

Thus $u_{x} = v_{y}$ and $u_{y} = - v_{x}$ at $0$ : both Cauchy–Riemann equations hold there.

step 1.2

2.2

Along nonzero real $z = t$ , the quotient in step 1.3 is $1$ . Along $z = t (1 + i)$ , it is $(1 - i)^{2} / (1 + i)^{2} = - 1$ . Both paths tend to $0$ , so [L1] shows that $f^{'} (0)$ does not exist.

step 1.3L1algebra

3.1

Steps 1.2 and 2.1 verify the false claim's entire hypothesis at $0$ , while step 2.2 denies its conclusion. This also exhibits the missing ingredient in [L2]: the coordinate map is not real totally differentiable at $0$ .

step 1.2step 2.1step 2.2L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Howell and Mathews, Complex Analysis, Example 3.2.5 (standard reference, not scraped)