2
$\begingroup$

I came across the following problem that says:

Consider the functions $f,g\colon \mathbb C \rightarrow \mathbb C$ defined by $f(z)=e^z, g(z)=e^{iz}$. Let $S=\{z \in \mathbb C:Re(z)\in [-\pi,\pi]\}.$ Then which of the following options is correct?

$1$. $f$ is an onto entire function

$2$. $g$ is a bounded function on $\mathbb C$

$3$. $f$ is bounded on $S$

$4$. $g$ is bounded on $S$

Can someone point me in the right direction? Thanks in advance for your time.

$\endgroup$
  • 2
    $\begingroup$ What does $[-z, z]$ mean? $\endgroup$ – Michael Albanese Feb 3 '13 at 3:12
  • $\begingroup$ I'm sure that you meant something else than $[-z,z]$, since $S$ makes no sense as it is. $\endgroup$ – Cameron Buie Feb 3 '13 at 3:36
  • $\begingroup$ sorry for the mistake. I have edited it. $\endgroup$ – user52976 Feb 3 '13 at 3:41
  • $\begingroup$ Thanks for remarking that point +1 $\endgroup$ – mrs Feb 4 '13 at 17:25
2
$\begingroup$

For the first statement, can you find $z \in \mathbb{C}$ such that $f(z) = 0$? For the rest, use the fact that $$|e^z| = |e^{a+bi}| = |e^ae^{bi}| = |e^a||e^{bi}| = |e^a| = e^a = e^{\Re(z)}.$$

$\endgroup$
  • $\begingroup$ Thanks a lot sir.I have got it. $\endgroup$ – user52976 Feb 3 '13 at 4:18
  • $\begingroup$ $f$ is bounded only $\endgroup$ – Marso May 3 '13 at 6:43
  • $\begingroup$ Why $e^{iz}$ is not bounded in $\mathbb{C}$ @LaBelleNoiseuse $\endgroup$ – Kushal Bhuyan Oct 24 '15 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy