Question

Let Ω ⊂ C be open, f : Ω → C holomorphic and z0 ∈ Ω. Suppose f(z0) 6= 0.

Show that there is an open neighborhood U ⊂ Ω of z0 and g : U → C holomorphic s.t.
eᵍ(z) = f(z), z ∈ U.

Answer 1

The answer explains how to show the existence of a holomorphic function g for which e^g(z) equals the given holomorphic function f(z), based on properties of the exponential function and branches of the logarithm.

Step-by-step explanation:

The question asks us to show that if Ω ⊂ C is open, f : Ω → C is holomorphic and z0 ∈ Ω such that f(z0) ≠ 0, then there is an open neighborhood U ⊂ Ω of z0 and a holomorphic function g : U → C such that e⁰(z) = f(z) for all z ∈ U. To demonstrate this, we use the fact that the exponential function is locally invertible in the complex plane, and since f(z0) is non-zero, we can choose a branch of the logarithm to define g such that g(z) is holomorphic on U and e⁰(z) will actually be equal to f(z). This is a consequence of the existence of the local inverse and the properties of holomorphic functions.