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Let $f$ be a non-constant holomorphic(analytic) function in the unit disc$\{|z|<1\}$ such that $f(0)=1$ then it is necessary that

$(1)$ there are infinitely many points $z$ in the unit disc such that $|f(z)|=1 $

$(2)\ f$ is bounded

$(3)$ there are at most finitely many points in the unit disc such that $|f(z)|=1$

$(4)$ $f$ is rational function.

I tried the if (2)is true then and it is given that $f$ analytic then it will be constant.Am I right? How to disprove of prove other option($\textbf{ Here answer is unique}$).please help me.thanks in advance.

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  • $\begingroup$ why do you think $2$ is false?? Liouville's?? check the statement of Liouville's... $\endgroup$
    – user87543
    Jan 30 '14 at 16:08
  • $\begingroup$ In the statement it is given on the complex plane. $\endgroup$ Jan 30 '14 at 16:11
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    $\begingroup$ If i take $f(z)=\frac{1}{z+1}$ then it is the counter example for my statement it is bounded.$f(0)=1$ also non constant but not constant on the complex as well as not analytic on the complex plane. $\endgroup$ Jan 30 '14 at 16:17
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    $\begingroup$ for fourth option how does exponential function behave? $\endgroup$
    – user87543
    Jan 30 '14 at 16:25
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    $\begingroup$ @SiddhantTrivedi $f(z)=e^z$; so you're left with either 1 or 3 $\endgroup$
    – egreg
    Jan 30 '14 at 17:11
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Hints:

  1. Note that (1),(3) are the negation of one another. This implies that for any give $f$, exactly one of them is the case (but, perhaps not that different examples cannot admit different cases).

  2. Restate the context: $f$ is analytic on $\Bbb D$ and $1\in S^1\cap\mathrm{Rng}(f)$. We're asked whether $|S^1\cap\mathrm{Rng}(f)|$ is finite or infinite.


Added I think it's safe to elaborate on my earlier hints. Truth be told, though, I haven't noticed that the answer should best rely on the maximum modulus principle. In particular, I based my remarks on the open mapping theorem (which the maximum modulus principle can be viewed as a special case of). Indeed, if the range of $f$ contains $1$, it also contains some neighborhood thereof, which intersects a segment of the unit circle $S^1$. Since all these points must have distinct preimages, $f$ maps to $S^1$ on infinitely (uncountably) many points of $\Bbb D$.

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  • $\begingroup$ Can you give me more hint for (1) and (3) $\endgroup$ Jan 30 '14 at 17:22
  • $\begingroup$ In due time, I will, but this is homework; I will say this: it matters less that $1\in S^1\cap\mathrm{Rng}(f)$ and more that $S^1\cap\mathrm{Rng}(f)\neq\emptyset$. $\endgroup$ Jan 30 '14 at 17:44
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EDIT: Fixed as per comment

To contradict (2), $f(z) = \frac{1}{1-z}$

To contradict (3), $f(z) = 2z$ maps the circle radius 0.5 onto the circle radius 1, so there will be infinitely many z with f(z) having magnitude 1

To contradict (4), $f(z) = exp(z)$ is not a rational function.

So it remains to prove (1)

Sketch proof
Take an infinite series of disks centered around $z=0$, of decreasing radius (starting at $r=0.9$ say) and show that each disk must contain some point z (not = 0) with $|f(z)| = 1$ (either on the boundary or in the interior). Then use $\frac{1}{2}|z|$ for the radius of the next disk...

Proof:
Maximum Modulus Theorem says that f will achieve its maximum somewhere on the boundary. So there will be some point $z_{max} s.t. |f(z_{max})| \ge 1$ on the boundary.

The Minimum Modulus Theorem gives a weaker result:

'If f is holomorphic and non-constant on a bounded domain D, then f attains its minimum either at a zero of f, or on the boundary.

So f must have a minimum $z_{min} s.t. |f(z_{min})| < 1$ somewhere in or on this disk. (NOTE: It can't be > 1 as $f(0)=1$, and if it were = 1, we would have z constant on the entire disk, which will probably break the premise (*)).

So tracing a line (taking care to avoid passing through the origin) from z_max to this z_min, (as f is continuous) there must exist some z s.t. $f(z)$ has magnitude 1.

So add that z to your set of solutions, pick a new disk with radius $\frac{1}{2}|z|$, rinse and repeat.

(*) It isn't yet rigourous as this has still to be proved...

PS: could the questioner please tidy up the question? It doesn't have a sensible title, it doesn't read properly (it should be clearly stating that exactly ONE of the following four statements are true), and the following paragraph is confused and doesn't make sense.

(EDIT: Note to self: https://math.stackexchange.com/help/notation , thanks Jonathan Y)

EDIT: onto Jonathan's challenge: Proving uncountably many points is a good challenge, I can see it is aesthetically a better question, as a stronger condition will elicit a deeper understanding. Hmm, |f(z)|=1 contours passing through z=0...

|f(z)|=1 means that f(z) is on the unit circle.

So maybe we could look at the inverse mapping; $f^{-1}(unit circle)$ i.e. what maps to the unit circle?

If f is holomorphic, this means it is continuous, which means the inverse map must also be continuous...

So perform the inverse map of the unit circle, and you must get some loop, by continuity. And we know this passes through 0. Which gives us uncountably many points in ANY neighbourhood of 0.

Much cleaner! Working to prove the stronger condition yielded a much simpler solution, good call Jonathan!

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  • $\begingroup$ Sometimes there more than one answer.please check my profile and question. $\endgroup$ Jan 31 '14 at 3:43
  • $\begingroup$ Nice. A small correction: we require the magnitude on the boundary to be strictly greater than 1 in order for a proper point to exist inside each disc. That's just fine, though, as the maximum modulus principle still works to give us that. BTW can you show that there are uncountably many such points? (Also, please recall that this is homework, so even a complete solution might be premature right now) $\endgroup$ Jan 31 '14 at 6:02
  • $\begingroup$ Actually it is perfectly okay if the |f(z)|=1 point lies ON the boundary. Note that I chose the first closed disk to have radius 0.9 not 1. Whether on or in, we get a nonzero z s.t.|f(z)|=1, with z (on or) in this closed disk and not = 0. Then we create a new disk with radius 0.9 |z|, and repeat the process. $\endgroup$
    – P i
    Jan 31 '14 at 8:03
  • $\begingroup$ Yes, you're correct. My mistake. $\endgroup$ Jan 31 '14 at 12:11
  • $\begingroup$ @P-i-: how do we know that $f$ is invertible? (In general, it needn't be.) Also, I added more detail in my answer below (which I now realize might be considered overkill in this context, although still of some interest), if you're interested. $\endgroup$ Feb 2 '14 at 21:59
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Consider a single function $$f(z)=\exp{\frac{z}{1-z}}.$$Then , clearly options $2,3,4$ are FALSE.

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