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Cartesian and polar forms of the Cauchy–Riemann equations agree away from the origin

Statement

Let $f = u + i v$ be real totally differentiable on an open subset of $C$ . On an open set of parameters $(r, θ)$ with $r > 0$ and $r e^{i θ}$ in the domain, put

$U (r, θ) = u (r \cos θ, r \sin θ), V (r, θ) = v (r \cos θ, r \sin θ) .$

At every such parameter point, the Cartesian Cauchy–Riemann equations are equivalent to

$U_{r} = \frac{1}{r} V_{θ}, V_{r} = - \frac{1}{r} U_{θ} .$

When these conditions hold,

$f^{'} (r e^{i θ}) = e^{- i θ} (U_{r} + i V_{r}) .$

No assertion is made at $r = 0$ , and no global choice of argument is used.

Facts & Assumptions

Given: A real totally differentiable $f = u + i v$ , a parameter point $(r, θ)$ with $r > 0$ , and the polar pullbacks $U, V$ stated above.

[L1]

The real total-derivative chain rule is $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[L2]

The real derivatives satisfy $(\sin x)^{'} = \cos x$ and $(\cos x)^{'} = - \sin x$ (The derivatives of sine and cosine are cosine and minus sine).

[F1]

For real $x, y$ , $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ ; in particular $e^{i θ} = \cos θ + i \sin θ$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[L3]

For a real-differentiable complex-valued map, complex differentiability is equivalent to the Cartesian Cauchy–Riemann equations, and then $f^{'} = u_{x} + i v_{x}$ (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).

Proof

technique · direct

1.1

Put $c = \cos θ$ and $s = \sin θ$ . By [L1] and [L2], $U_{r} = u_{x} c + u_{y} s$ and $U_{θ} = r (- u_{x} s + u_{y} c)$ .

givenL1L2

1.2

The same calculation gives $V_{r} = v_{x} c + v_{y} s$ and $V_{θ} = r (- v_{x} s + v_{y} c)$ .

givenL1L2

2.1

If $u_{x} = v_{y}$ and $u_{y} = - v_{x}$ , steps 1.1–1.2 give $V_{θ} = r U_{r}$ and $U_{θ} = - r V_{r}$ , which are the polar equations because $r > 0$ .

step 1.1step 1.2algebra

2.2

Conversely, the inverse coordinate formulas are $u_{x} = U_{r} c - (U_{θ} / r) s$ , $u_{y} = U_{r} s + (U_{θ} / r) c$ , $v_{x} = V_{r} c - (V_{θ} / r) s$ , and $v_{y} = V_{r} s + (V_{θ} / r) c$ . Substituting the polar equations gives $u_{x} = v_{y}$ and $u_{y} = - v_{x}$ .

step 1.1step 1.2givenalgebra

3.1

Under either equivalent form, [L3] and step 2.2 give $f^{'} = u_{x} + i v_{x} = (c - i s) (U_{r} + i V_{r}) = e^{- i θ} (U_{r} + i V_{r})$ by [F1].

step 2.2L3F1algebra∎

Depends on

Used by

The complex exponential satisfies the Cauchy–Riemann equations in Cartesian and polar form Example

Dependency tree · two levels

23 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

J. Lebl, Guide to Cultivating Complex Analysis, Exercise 2.1.8(a) (standard reference, not scraped)
R. Howell and J. Mathews, Complex Analysis, Theorem 3.2.10 (standard reference, not scraped)