# Complex analysis question.

Which of the following are not true?

$(a)$ There exists an analytic function $f:\mathbb{C}\to\mathbb{C}$ such that for all $z\in \mathbb{C}$ , $Re(f(z))=e^x$.

$(b)$ There exists an analytic function $f:\mathbb{C}\to\mathbb{C}$ such that $f(0)=1$ and for all $z\in \mathbb{C}$ such that $|z|\geq1$ such that $|f(z)|\leq e^{-|z|}$.

could you please give me some hints?

I have applied liouville thorem for $(b)$ and got $f(z)$ is constant but not necessarily $f(z)=1$. Am I correct?

• Well, if $f$ is constant and $f(0)=1$, then ... But does that comply with $|f(z)|\le e^{-|z|}$? Or more directly: Use Cauchy integral to compute $f(0)$ and compare with the given $f(0)=1$. – Hagen von Eitzen May 8 '17 at 6:18
• Yes. Could you please give me some hints? – user444042 May 8 '17 at 6:19

(a): If $u=Re(f)$, then $u$ is harmonic, this means: $u_{xx}+u_{yy}=0$. Is this possible if $u(x,y)=e^x$ ?

(b) For $|z| \ge 1$ we have $e^{-|z|} \le 1/e$, hence $f$ is bounded for $|z| \ge 1$.

The set $K=\{z :|z| \le 1\}$ is compact, hence $f$ is bounded on $K$.

Conclusion: $f$ is bounded on $\mathbb C$, hence $f$ is constant, thus, for some $c$:

$f(z)=c$ for all $z$. From $f(0)=1$ we get $c=1$. It follows:

$1=f(1) \le 1/e$, a contradiction.

• Awesome. Thank you very much. – user444042 May 8 '17 at 12:32