Question

f is a nonconstant complex function (the domain is the complex numbers and the range is the complex numbers). f is entire (it is differentiable everywhere).

Let S be an open set in the complex plane. Show that there exists a z such that f(z) is inside S.

Answer 1

@satellite73 @TuringTest help?

Answer 2

look at it this way, if f(z) has a real derivative at every point, then f is continuous throughout the complex plane. that means that if S is an open set of the complex plane, then there must be some point z where f(z) exist inside that set, right?

Answer 3

It's a good idea, but continuity alone does not imply that we cover all of C (the complex numbers) in our range.

Answer 4

@Eureka70 are you familiar with the following theorem? If f is entire and nonconstant (and the domain is the complex plane), then the range of f is not bounded.

Answer 5

no i was never taught complex numbers formally, learned what i know myself, but doesn't that give you your answer? if the range of f is not bdd, then at least one value of f(z) should be inside S if S is an open set

Answer 6

Not necessarily. The range is two dimensional. Suppose S is all complex numbers with a negative imaginary part (the lower half of the plane). Suppose the range of f is the upper half of the plane. Then the range is unbounded because it goes upwards infinitely, but it does not intersect S.

Answer 7

yea, but then i have no idea how it is that you are supposed to show that the open set and the range of f will ever meet if you get what i'm thinking

Answer 8

This is meant to be a difficult problem. I can't seem to get any ground on it. I tried a number of things. I tried letting g(z) = s + 1/f(z), where s is a point in S. Then I wanted to say that because f is entire, it is unbounded, and therefore g will get very close to s and therefore intersect S.

But I wasn't sure where to go from there. @Eureka70 @satellite73 @TuringTest

Answer 9

I think I have a solution. Assume that f does not intersect S. Then g = 1/(f-s) is defined for all complex numbers and its entire.

But by our theorem, this means that |g| is unbounded. So if |1/(f-s)| can get as large as we want, it means that f can get as close to s as we want. So f must enter into S to do this.

This is a contradiction, so the assumption that f does not intersect S is incorrect.

Answer 10

Thoughts?