Does There Exist an Analytic Function?

In complex analysis, the study of analytic functions (or holomorphic functions) is a fundamental area. This article aims to explore a specific problem: the existence of an analytic function f(z) such that it maps the closed disk (|z| leq 1) to a desired configuration while satisfying certain conditions at the boundary and within the disk.

The problem at hand is to determine if there exists an analytic function on the disk (|z| leq 1) that meets the following criteria:

At the origin, (f(0) i) At the boundary, (f(z^2) z) for (|z| 1) The function has a zero at (z_1)

Two examples of such functions are provided:

Example 1: A Specific Function f(z)

The first function given is:

$$f(z) 21i cdot frac{1-2z}{2-z}$$

f(z)  21i cdot frac{1 - 2z}{2 - z}

Analyticity: This function is analytic on the disk (|z| leq 1) because the denominator (2 - z eq 0) for (|z| leq 1) and z 2 is outside the disk. Value at Zero: At (z 0), the function value is indeed (f(0) 21i cdot frac{1 - 0}{2 - 0} i). Zero Location: The zero occurs at (z frac{1}{2}), which is inside the disk (|z| leq 1). Boundary Condition: At the boundary (|z| 1), we need to verify (f(z^2) z).

Verification on the Boundary

On the boundary, where (|z| 1), the transformation is:

$$frac{1 - 2z}{2 - z}$$ Let's verify the boundary condition: Consider (z e^{itheta}) for (0 leq theta $frac{1 - 2z}{2 - z}$ can be rewritten as: $$frac{1 - 2e^{itheta}}{2 - e^{itheta}}$$ Multiply by the complex conjugate: $$frac{(1 - 2e^{itheta})(2 - e^{-itheta})}{(2 - e^{itheta})(2 - e^{-itheta})}$$ Simplify the numerator: $$frac{(1 - 2e^{itheta})(2 - e^{-itheta}) (2 - e^{-itheta} - 4e^{itheta} 2e^{itheta}e^{-itheta}) 2 - 2e^{-itheta} - 2e^{itheta} 2$$ The denominator: $$frac{(2 - e^{itheta})(2 - e^{-itheta}) 4 - 2e^{-itheta} - 2e^{itheta} 1 5 - 2(e^{itheta} e^{-itheta}) 5 - 4cos(theta)$$ Since (z^2 e^{2itheta}), we have: $$frac{2(1 - 2e^{itheta})}{5 - 4cos(theta)} e^{itheta}$$ This confirms (f(z^2) z).

Example 2: A General Function f(z)

A general form of the function is given by:

$$f(z) frac{1 - i}{alpha} cdot frac{alpha - z}{1 - overline{alpha}z}$$

f(z)  frac{1 - i}{alpha} cdot frac{alpha - z}{1 - overline{alpha}z}

Here, α is a constant satisfying (|alpha| 1). This form ensures that (frac{alpha - z}{1 - overline{alpha}z}) maps the unit disk onto itself.

Construction Explanation

The crucial point is that (frac{alpha - z}{1 - overline{alpha}z} 1) when (z alpha). The example used in the reference solution is with (alpha frac{1}{2}).

In summary, both the specific and general forms of the function f(z) demonstrate the existence of analytic functions that satisfy the given conditions, providing a valuable example for understanding complex analysis and conformal mapping.