# Thread: Complex analysis, accumulation points

1. ## Complex analysis, accumulation points

Suppose that a function $f$ that is analytic in some arbitrary region Ω in the complex plane containing the interval [1,1.2]. Assume $f(x) = \cot (x)$ for all $x \in [1,1.2]$. Show that $f(z) = -i$ has no solutions in Ω.
So I understand how to show that there is no solutions, but my lecturer said before that, I must show that $f(z) = \cot (z)$ for all $z \in \Omega$

I'm not sure how to explain it. There are some theorems that say if $f(z) = ...$ (for z on some interval) has an accumulation point a the origin then $f(z) = ...$ for all z in the complex plane, but this equation has no accumulation point at the origin.

So how do I go about showing f(z) = cot(z) for all z, instead of just the small interval given?

2. ## Re: Complex analysis, accumulation points

Are you sure you're not being asked to show that f(z) = cot(z) is ANALYTIC for all z? f can be ANY function (by definition, a function is defined by the user)...

3. ## Re: Complex analysis, accumulation points

Originally Posted by Prove It
Are you sure you're not being asked to show that f(z) = cot(z) is ANALYTIC for all z? f can be ANY function (by definition, a function is defined by the user)...
Nope, the question was copy/pasted exactly as is, with a few LaTex added in.

At first I was thinking since f is analytic then it can be represented as a power series. And by the uniqueness theorem for the power series, I can argue that it must be cot(z) everywhere, but the only thing missing is the accumulation point at the origin.

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