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I was thinking about the problem that says:

Let $f \colon \mathbb C \rightarrow \mathbb C$ be defined by $f(z)=\cos z.$ Then which of the following options is correct?
$1.|f(z)| \leq 1$
$2.|f(z)| \leq \pi$
$3.|f(z)| \leq |z|$
$4. f$ is unbounded.

My Attempt: option $1$ is clearly false since $\cos i=f(i)=1+\frac{1}{2!}+\frac{1}{4!}+\cdots >1.$ But I am not sure about the other options. Another observation: Also,I see $f(z)$ is analytic on $\mathbb C$ and it is non-constant. So ,It must be unbounded( by Liouville's theorem).So option $4.$ looks right.Am I right ?

Thanks in advance for your time.

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    $\begingroup$ You know what $\cos i$ is; so what about $\cos 2i$, $\cos 3i$, etc.? $\endgroup$ – Jonas Meyer Feb 10 '13 at 5:01
  • $\begingroup$ Sir,I see $f(z)$ is analytic on $\mathbb C$ and it is non-constant. So ,It must be unbounded( by Liouville's theorem).So option $4.$ looks right.Am I right ? $\endgroup$ – user52976 Feb 10 '13 at 5:06
  • $\begingroup$ Also, you can try plugging $\:z=0\:$ into option 3. $\;\;$ $\endgroup$ – user57159 Feb 10 '13 at 6:32
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$4.\cos z$ is unbounded.

We know that $\cos (iy)=\cosh y$ and $\sin(iy)=i \sinh y $ $$\cos z=\cos x \cosh y-i \sin x \sinh y$$ so, we get $$|\cos z|^2=\cos^2 x+ \sinh^2 y$$ now it is easy to see that $\cos z$ is unbounded.Can u see it?

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  • $\begingroup$ Yes.I have got it. Also,we can apply Lioville's theorem to reach the same conclusion. Am I right?(Please see my Attempts). $\endgroup$ – user52976 Feb 10 '13 at 5:17
  • $\begingroup$ I don't know about the Lioville's theorem but this is a simplest way to conclude that $\cos z$ is unbounded. $\endgroup$ – Siddhant Trivedi Feb 10 '13 at 5:23
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There is a high-tech way to do this (Liouville's theorem). But there is a lo-tech way.

  1. You've already done this.

  2. You've note that $\cos(ix)=\cosh(x)$ which is unbounded on $\mathbb{R}$.

  3. How about $z=0$?

So..

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