iamqsqsqs said:Difficult with this problem.
Find all entire functions f such that |f(z)| = 1 for all z with |z| = 1.
Are there any vigorous proofs?
DonAntonio said:By the minimum modulus theorem, such a function must be constant...
DonAntonio
micromass said:So the function f(z)=z is a constant then??
An entire function is a complex-valued function that is defined and analytic (meaning it has a derivative at every point in its domain) over the entire complex plane.
This can be proved using the maximum modulus principle, which states that the maximum value of a complex-valued function on a closed bounded region occurs on the boundary of that region. Therefore, if we can show that |f(z)|=1 on the boundary of the unit circle, then it must be true for all points within the unit circle as well.
This means that the absolute value (or modulus) of the function is equal to 1 for all points on the unit circle. In other words, the function stays on the unit circle and does not extend beyond it in any direction.
To prove that a function is entire, we must show that it is analytic (has a derivative at every point in its domain) and defined over the entire complex plane. This can be done using the Cauchy-Riemann equations, which state that a function is analytic if its partial derivatives satisfy certain conditions. We must also show that the function is continuous on the complex plane, as this is a necessary condition for analyticity.
Yes, there are other methods such as using the Cauchy integral formula, which relates the value of a function at a point to its values on the boundary of a region. Another method is using the Taylor series expansion of the function, which shows that the function can be represented as an infinite sum of powers of the variable z. Additionally, the Laurent series can also be used to prove that a function is entire by showing that it has no singularities (or poles) in its domain.