Bounded complex valued function

In summary, the conversation discusses the problem of proving that a function f(z) is equal to a constant c if it is analytic on the entire complex plane and satisfies certain conditions regarding its modulus. The first statement uses the Maximum Modulus Theorem and Liouville's Theorem to prove this, while the second statement requires a more complex approach involving the derivative of f at z=0. Ultimately, the conclusion is that f(z) must be equal to zero, rather than just a constant, in the second scenario.
  • #1
g1990
36
0

Homework Statement


1. f(z) is a function that is analytic on all of the complex plane, and mod(f)<=mod(z). Prove that f=cz.
2. f(z) is analytic on all of the complex plane, and mod(f)<= sqrt(mod(z)). Prove that f is constant


Homework Equations


Liouvilles thm: the only bounded entire functions are constant
Maximum modulus thm: If g is analytic on a domain D and achieves a maximum modulus on the interior of D, then g is constant


The Attempt at a Solution


1. I think I have this one. So, |f(z)/z|<=1, and analytic enerywhere except at z=0. If z=0 is an essential singularity or a pole, f(z)/z is not bounded near z, which is a contradiction. Thus, z=0 is a removable singularity or f(z)/z. So, There is an entire function h(z) where h(z)=f(z)/z when z is not 0, and h(z)=c when z=0 for some value c. Then, |h(z)| is bounded by max(1,c), and is constant. Then, f(z)/z=c for all z not equal to zero, so that f=cz. At zero, f(z)=0 by continuity because f is analytic everywhere on C. thus, f(z)=cz on the complex plane.
2. I don't know where to begin here. I would like to show that f is bounded, and then use Liouville's thm, but I don't know if that's the way to go.
 
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  • #2
Start from |f(z)^2/z|<=1.
 
  • #3
In case of 2), you can use that the inequality implies that f(0) = 0. Then the fact that f is analytic and thus complex differentiable at zero implies that the limit of f(z)/z for z to zero must exist as it is the derivative of f at z = 0. Then you can complete the proof by splitting the complex plane in some small neighborhood of zero and the rest of the complex plane and considering the behavior of f(z)/z.

You will be led to a stronger conclusion than asked in the problem: f(z) is actually zero.
 

FAQ: Bounded complex valued function

What is a bounded complex valued function?

A bounded complex valued function is a function that maps complex numbers to complex numbers and has a finite limit at every point in its domain. In other words, the function does not have any infinite or undefined values and its values do not grow beyond a certain limit.

What are some examples of bounded complex valued functions?

Examples of bounded complex valued functions include polynomials, trigonometric functions, and exponential functions. For instance, the function f(z) = z^2 is a bounded complex valued function since its values are always finite and do not grow beyond a certain limit.

How is the boundedness of a complex valued function determined?

The boundedness of a complex valued function can be determined by evaluating its limit at every point in its domain. If the limit exists and is finite at every point, then the function is bounded. Additionally, the absolute value of the function's values should not exceed a certain limit.

What is the significance of bounded complex valued functions in mathematics?

Bounded complex valued functions are important in mathematics because they have well-defined properties and behaviors. They are used in various mathematical theories and applications, such as complex analysis, differential equations, and signal processing.

How does the boundedness of a complex valued function affect its continuity?

A bounded complex valued function is always continuous, meaning that it has no sudden jumps or breaks in its values. This is because the function's values do not grow beyond a certain limit, allowing for a smooth and continuous mapping from its domain to its range.

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