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

Example

The complex exponential is entire with derivative itself. Its Cartesian components satisfy the Cartesian Cauchy–Riemann equations everywhere, and its polar components satisfy the polar equations at every parameter point with $r > 0$ .

Facts & Assumptions

Given: Complex numbers $z, h$ , Cartesian coordinates $z = x + i y$ , and polar parameters $z = r e^{i θ}$ with $r > 0$ when polar coordinates are used.

[F1]

The complex exponential is defined by $\exp z = \sum_{n \geq 0} z^{n} / n!$ (The complex exponential by its power series), and this series converges absolutely for every complex $z$ (The complex exponential series converges absolutely for every complex argument).

[L2]

For all complex $z, w$ , $\exp (z + w) = \exp z \exp w$ , and the complex exponential restricts to the real exponential on the real axis ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

[L3]

$\exp (x + i y) = e^{x} (\cos y + i \sin y)$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[L4]

The complex exponential is entire and has derivative $\exp^{'} (z) = \exp z$ (The complex exponential is entire and its complex derivative is itself).

[L5]

Away from $r = 0$ , the polar Cauchy–Riemann equations are $U_{r} = r^{- 1} V_{θ}$ and $V_{r} = - r^{- 1} U_{θ}$ (Cartesian and polar forms of the Cauchy–Riemann equations agree away from the origin).

Verification

technique · direct computation

1.1

For $h \neq 0$ , the addition law and defining series give $\frac{e^{z + h} - e^{z}}{h} = e^{z} \frac{e^{h} - 1}{h} = e^{z} (1 + \sum_{n \geq 2} \frac{h^{n - 1}}{n!}) .$

F1L2algebra

1.2

From [L3], $u = e^{x} \cos y$ and $v = e^{x} \sin y$ . Therefore $u_{x} = e^{x} \cos y = v_{y}$ and $u_{y} = - e^{x} \sin y = - v_{x}$ , and $u_{x} + i v_{x} = e^{x + i y}$ .

L3algebra

1.3

Put $c : = \cos θ$ , $s : = \sin θ$ , $A : = e^{r c}$ , and $ϕ : = r s$ . Then $U = A \cos ϕ$ and $V = A \sin ϕ$ , so $U_{r} = A (c \cos ϕ - s \sin ϕ), V_{r} = A (c \sin ϕ + s \cos ϕ),$ $U_{θ} = - r V_{r}, V_{θ} = r U_{r} .$

L3algebra

2.1

When $0 < ∣ h ∣ \leq 1$ , absolute convergence at $1$ gives the finite constant $C : = \sum_{n \geq 2} 1 / n!$ , and $∣ \sum_{n \geq 2} \frac{h^{n - 1}}{n!} ∣ \leq ∣ h ∣ \sum_{n \geq 2} \frac{1}{n!} = C ∣ h ∣ ⟶ 0.$

step 1.1F1algebra

3.1

Thus the quotient tends to $e^{z}$ , directly confirming $\exp^{'} (z) = \exp z$ and [L4].

step 1.1step 2.1L4

4.1

Since $r > 0$ , step 1.3 is equivalent to $U_{r} = r^{- 1} V_{θ}$ and $V_{r} = - r^{- 1} U_{θ}$ , exactly [L5]. At $r = 0$ these polar identities are not asserted; the Cartesian calculation in step 1.2 covers the origin.

step 1.2step 1.3L5∎

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Sources

J. Lebl, Guide to Cultivating Complex Analysis (standard reference, not scraped)
MIT 18.04, Topic 2: Functions of a Complex Variable (standard reference, not scraped)