CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-02

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$f (x + i y) = e^{x} (\cos 2 y + i \sin 2 y)$ is continuous, satisfies $f (z + w) = f (z) f (w)$ and $f (1) = e$ , but is not the standard complex exponential

Statement refuted

Continuity, $f (z + w) = f (z) f (w)$ , and $f (1) = e$ do not characterize the standard complex exponential. The conventions and prerequisite facts used below are recorded in $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ , The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane, The addition formulas for sine and cosine, The derivatives of sine and cosine are cosine and minus sine, Quarter-turn values and shifts by pi/2 and pi, The exponential function is strictly increasing, A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions.

Facts & Assumptions

Given: $f (x + i y) = e^{x} (\cos 2 y + i \sin 2 y)$ .

[L1]

The addition formulas for sine and cosine gives the sine and cosine formulas for every pair of real arguments.

[L2]

The exponential function is strictly increasing states that $x \mapsto e^{x}$ is continuous, and The derivatives of sine and cosine are cosine and minus sine makes sine and cosine continuous.

[L3]

A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions makes a complex-valued map continuous when its real and imaginary parts are continuous.

[L4]

$\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential restricts to the real exponential law $e^{x + s} = e^{x} e^{s}$ .

[L5]

Quarter-turn values and shifts by pi/2 and pi gives $\sin (2 π) = 0$ and $\cos (2 π) = 1$ .

[L6]

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

Counterexample

1.1

For $z = x + i y$ and $w = s + i t$ , [L4] and [L1] give $f (z + w) = e^{x + s} (\cos (2 y + 2 t) + i \sin (2 y + 2 t)) = f (z) f (w)$ , and $f (1) = e$ .

L1L4algebra

1.2

The coordinate maps $(x, y) \mapsto x$ and $(x, y) \mapsto 2 y$ are continuous by the Euclidean norm estimate. Composition with the continuous real functions in [L2] is continuous, and the identity $u v - u_{0} v_{0} = u (v - v_{0}) + v_{0} (u - u_{0})$ proves continuity of their products. Thus [L3] makes $f$ continuous.

L2L3

2.1

By [L5], $f (i π) = 1$ , while [L6] gives $\exp (i π) = - 1$ . Thus $f$ is not the standard exponential.

L5L6∎

Depends on

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Sources

J. Lebl, Basic Analysis I: Complex Numbers and the Complex Exponential (standard reference, not scraped)

f(x+iy)=ex(cos⁡2y+isin⁡2y)f(x+iy)=ex(cos2y+isin2y) is continuous, satisfies f(z+w)=f(z)f(w)f(z+w)=f(z)f(w) and f(1)=ef(1)=e, but is not the standard complex exponential