Find all analytical functions with this property

Question

This is one of my homework questions:

Is there any function $f$ which is analytical on open unit disk and in this region we have

$$\left | f(z) \right | = e^{\left | z \right |}$$

My attempt to solve question:

Since $f$ is analytical, so $ln(f(z))$ and $\measuredangle f(z)$ are harmonic conjugate, so they are satisfy following Laplace equation:

$u_{xx}+u_{yy}=0$

this leads to $$\frac{1}{\sqrt{x^2+y^2}}=0$$

Which is contradiction, so there is no function with this property,

However, as you know, $Ln(z)$ is not analytical on $\Im(z)=0 \cap \Re(z)<0$ and this conclusion is not right in this way

Do you have any other Idea!?

Request for reference

Also, tell me please, in which complex analysis book I can find this kind of problems... I'm looking for some proving problems in complex analysis like this above.

Answer 1

One way to solve it:

Let $0<r<1$. By the Maximum Modulus Theorem, $|f(z)|\leq e^r$ for all $z\in \overline{B(0,r)}$ and as $f(z)$ never vanishes (because $|f(z)|=e^{|z|}$) we can apply MMT to $\frac{1}{f(z)}$ to obtain $|f(z)|\geq e^r$ for $z\in\overline{B(0,r)}$. Therefore $f$ is constant on $B(0,r)$ and by the Identity Theorem $f$ must be constant on all of the unit disk, which contradicts that $|f(z)|=e^{|z|}$ so a function with that property cannot exist.

Answer 2

You were near to a correct proof!

If $f$ is analytic in $D$ and $|f(z)|=e^{|z|}$ then $f(z)\ne0$ in $D$. As $D$ is simply connected this implies that there is a logarithm of $f$ in $D$, i.e., an analytic function $g: \>D\to {\mathbb C}$ with $$e^{g(z)}=f(z)\qquad(z\in D)\ .$$ We then have $$e^{{\rm Re}\bigl(g(z)\bigr)}=|f(z)|=e^{|z|}\qquad(z\in D)$$ and therefore ${\rm Re}\bigl(g(z)\bigr)\equiv|z|$. This is impossible since the LHS is a harmonic function and the RHS is not.