Is the complex space with the dscnf metric complete?

Question

Let $d_{scnf}$ be a metric on the complex plane where, if $x=r_x$$e^{i\theta_x}$ and $y=r_y$$e^{i\theta_y}$ with $\theta_x$,$\theta_y$ $\in [0,2\pi$[ : $$d_{scnf}(x,y)= \begin{cases} \lvert r_x-r_y\rvert & \text{if $\theta_x$=$\theta_y$}\\ r_x+r_y & \text{if $\theta_x$$\ne$$\theta_y$ }\\ \end{cases} $$

is the complex space with the dscnf metric complete ?

I am trying to take a Cauchy sequence however I fail to prove anything because i have two cases of distance. Any idea if this space is complete ?

Answer 1

For notational simplicity, let $d=d_{scnf}$.

First, we notice that for $x=0$ there is no ambiguity (the argument of zero is undefined) since under each definition $d(z,0)=|z|$

Let $z_n$ be a Cauchy sequence in the metric $d$. Wlog we can assume $z_n \ne 0$ for infinitely many $n$ (otherwise trivially $z_n=0 \to 0, n \ge N$); we then have two cases:

1 Assume $\arg z_n, z_n \ne 0, n \ge N$ is non-constant for all $N>0$; in other words for any $N>0$ we can find $p>q \ge N, z_pz_q \ne 0, \arg z_p \ne \arg z_q$.

Then fixing $\epsilon >0$ and letting $N(\epsilon)$ st $d(z_m, z_n) < \epsilon, m,n \ge N(\epsilon)$, then for all non zero $z_n, n \ge N(\epsilon)$ and $p,q$ as above for $N=N(\epsilon)$ at least one of $\arg z_p, \arg z_q$ is different from $\arg z_n$, so wlog letting that index be $p$, we have $|z_n|+|z_p|=d(z_n, z_p) < \epsilon$ hence $|z_n| < \epsilon, N \ge N(\epsilon)$ (since that is automatically satisfied for the zero terms), so $z_n \to 0$ in the metric.

2. Assume $\arg z_n, z_n \ne 0$ is constant $\theta$ for $n \ge N_0$; then by the definition of $d$ we have that $d(z_n, z_m)=||z_n|-|z_m||, n,m \ge N_0$ (as the result also holds if any of $z_n, z_m$ is zero by the discussion above) and the result follows from the real line completeness as $|z_n|$ is then Cauchy in the real line usual distance, hence converges to some non negative $r$; if $r=0$ then $z_n \to 0$ clearly in $d$, while if $r \ne 0$ we have that $z_n \to re^{i\theta}$.

So indeed $d$ is complete, while it is far from being equivalent with the usual Euclidean metric