- #1
iamqsqsqs
- 9
- 0
Does there exist a holomorphic function f(z) on the unit disc and satisfies f(1/n) = f(-1/n) = 1/n^3 for every n in N?
micromass said:There does not even exist a continuous function that does this.
Citan Uzuki said:Last time I checked [itex]z \mapsto |z|[/itex] was continuous...
*** Last time I checked [itex]\,\,\displaystyle{\left|\frac{1}{n}\right|\neq \frac{1}{n^3}}[/itex] ...
DonAntonio ***
iamqsqsqs, try looking at the zeros of f(z) - z^3. Do they form an isolated set of points?
Citan Uzuki said:Sorry, brain fart. I meant to say [itex]z \mapsto |z|^3[/itex]
A holomorphic function is a complex-valued function that is differentiable at every point in its domain. It can also be described as a function that is analytic, meaning it can be represented by a convergent power series.
The unit disc is the set of all complex numbers with a distance of less than or equal to 1 from the origin on the complex plane. It can be represented by the equation |z| ≤ 1.
A holomorphic function on the unit disc is a function that is differentiable at every point within the unit disc, whereas a holomorphic function on the complex plane is a function that is differentiable at every point in the entire complex plane. This means that a holomorphic function on the unit disc has a smaller domain of differentiability.
Yes, a holomorphic function on the unit disc can have singularities, but only at the boundary of the disc. This is because the unit disc is a compact set, meaning it is closed and bounded, and thus the function must be continuous on its boundary.
Holomorphic functions on the unit disc are important in the study of conformal mapping, which is a type of transformation that preserves angles. This is because holomorphic functions on the unit disc are also conformal, meaning they preserve angles locally. This property makes them useful in mapping regions of the complex plane to the unit disc and vice versa.