TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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The complex exponential is entire and its complex derivative is itself

Statement

The complex exponential is entire, and

$\exp^{'} (z) = \exp z$

for every $z \in C$ .

Facts & Assumptions

Given: A complex number $z = x + i y$ and the published complex exponential.

[F1]

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

[L1]

The real exponential is $C^{\infty}$ and $(e^{x})^{'} = e^{x}$ (The exponential function is smooth and $(\exp)^{'} = \exp$ ).

[L2]

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

[L3]

A real function differentiable at a point is continuous there (A function differentiable at $c$ is continuous at $c$ ).

[L4]

Finite sums and products of continuous real-valued maps on a topological space are continuous (Sums, products, absolute values, finite maxima and minima, and quotients of continuous real-valued maps on a topological space are continuous where defined).

[L5]

Continuous first partial derivatives satisfying the Cauchy–Riemann equations give complex differentiability, and holomorphy when this holds at every point (Continuous first partial derivatives and the Cauchy–Riemann equations imply complex differentiability pointwise, and holomorphy when they hold throughout an open set). Where $f$ is complex differentiable, $f^{'} = u_{x} + i v_{x}$ (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

By [F1], the real and imaginary components are $u (x, y) = e^{x} \cos y$ and $v (x, y) = e^{x} \sin y$ .

givenF1

2.1

By [L1] and [L2], $u_{x} = e^{x} \cos y, u_{y} = - e^{x} \sin y, v_{x} = e^{x} \sin y, v_{y} = e^{x} \cos y .$

step 1.1L1L2algebra

3.1

The one-variable factors in step 2.1 are continuous by [L1]–[L3]; their pullbacks along the coordinate projections are continuous, and [L4] makes all four displayed partials continuous on $R^{2}$ .

step 2.1L1L2L3L4

4.1

Step 2.1 gives $u_{x} = v_{y}$ and $u_{y} = - v_{x}$ everywhere. By [L5], the complex exponential is entire and its derivative is $u_{x} + i v_{x} = e^{x} (\cos y + i \sin y) = \exp z$ .

step 1.1step 2.1step 3.1L5∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 145 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Lebl, Guide to Cultivating Complex Analysis, Exercise 2.1.4 (standard reference, not scraped)
J. Orloff, MIT 18.04 Topic 2, Example 2.11 (standard reference, not scraped)