Is the following statement true? f′(z) ≠ 0 everywhere implies f is injective

No. f(z) = eᶻ on ℂ f′ = eᶻ never vanishes, yet f(z) = f(z + 2πi). Non-vanishing derivative gives only local injectivity.

Counterexample

Is this true?

f′(z) ≠ 0 everywhere implies f is injective

No — it is false.

The counterexample

f(z) = eᶻ on $\mathbb{C}$

Why it fails

f′ = eᶻ never vanishes, yet $f(z) = f(z + 2\pi{}i)$ . Non-vanishing derivative gives only local injectivity.

Keep practicing

Execution slip

The idea was right. The computation was not.

Published 23 Aug 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.