Form of analytic function given its real part is a function of |z| What is the general form of $f(z)$ given its real part only depends on $|z|$
Here is the beginning of my solution in which I get stuck:
We can write $f(z) = u(x,y) + i v(x,y)$ on which the C-R equations hold.
We can be more precise and write: $u(x,y) = u(\sqrt{x^2 + y^2})$. We can therefore get a more precise expression of our C-R equations for this case:
$$\frac{\partial u(\sqrt(x^2+y^2))}{\partial x} = u'(\sqrt{x^2+y^2})(x^2+y^2)^{-1/2}x = \frac{\partial v}{\partial y}$$
and similarily:
$$\frac{\partial u(\sqrt(x^2+y^2))}{\partial y} = u'(\sqrt{x^2+y^2})(x^2+y^2)^{-1/2}y = \frac{- \partial v}{\partial x}$$
I then try to solve for $v$ by integrating with respect to $x$ and $y$ respectively but do not see how this would have a solution. From the above, we can also directly get: 
$$-x\frac{\partial v}{\partial x} = y \frac{\partial v}{\partial y}$$
How should I go about working through this question? Maybe expressing the modulo with complex conjugates? 
Any hints are most appreciated!
 A: We have that the real part of $f(z)$ depends solely on $\vert z \vert = r$; thus
$f(z) = u(z) + iv(z) = u(r) + iv(r,  \theta); \tag 1$
the Cauchy-Riemann equations in polar coordinates are
$u_r = \dfrac{1}{r}v_\theta, \tag 2$
$v_r = -\dfrac{1}{r}u_\theta; \tag 3$
from (3) we find, since $u_\theta = 0$,
$v_r = 0, \tag 4$
and thus
$v_{\theta r} = v_{r \theta} = 0; \tag 5$
if we differentiate (2) with respect to $r$, and then substitute (2), (5) in the result, we obtain
$u_{rr} = -\dfrac{1}{r^2}v_\theta + \dfrac{1}{r}v_{\theta r} = -\dfrac{1}{r}u_r, \tag 6$
whence
$(\ln u_r))_r = \dfrac{u_{rr}}{u_r} = -\dfrac{1}{r}; \tag 7$
integrating with respect to $r$:
$\ln u_r = -\ln r + \ln C = \ln (Cr^{-1}), \; C \in \Bbb R; \tag 8$
$u_r = Cr^{-1}; \tag 9$
$u(r) = C\ln r + D, \; D \in \Bbb R; \tag{10}$
combining (2) and (9),
$v_\theta = ru_r = C; \tag{11}$
$v = C\theta + E, \; E \in \Bbb R; \tag{12}$
$f(z) = u + iv = C\ln r + D + iC\theta + iE = C(\ln r + i\theta) + (D + iE); \tag{13}$
with $z = re^{i \theta}$, $D + iE = \alpha \in \Bbb C$,
$f(z) = C \ln z + \alpha. \tag{14}$
