Download Behavior of exp(1/z) in a neighborhood of 0

Behavior of exp(1/z) in a neighborhood of 0
Fact: Given any complex number ζ 6= 0 and any ε > 0, there are infinitely many z ∈ C with
|z| < ε such that exp(1/z) = ζ.
Proof: Let ζ 6= 0 be given, then there are real numbers r > 0 and −π < θ ≤ π such that
ζ = reiθ . Furthermore, for any integer k, ζ = rei(θ+2kπ) .
Now let ε > 0 be given, let k be an integer such that k > 1/ε, and set
ωk = ln r + i(θ + 2kπ).
Then on the one hand
exp(ωk ) = eln r ei(θ+2kπ) = rei(θ+2kπ) = ζ,
and on the other hand
p
|ωk | = (ln r)2 + (ϑ + 2kπ)2 ≥ |ϑ + 2kπ| > (2k − 1)π ≥ kπ > 1/ε,
(with room to spare). Therefore, if zk = 1/ωk , then |zk | < ε and
exp(1/zk ) = exp(ω) = ζ.
Since there are infinitely many integers k greater than 1/ε, there are infinitely many zk with
modulus less than ε satisfying exp(z) = ζ.
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