# In a complex metrizable topological vector space, $d(0,\alpha x)\neq |\alpha|d(0,x), \ \alpha \in \mathbb C.$

Let $$(X,\tau)$$ be a complex metrizable topological vector space with the metric $$d$$. Does the following hold: $$d(0,\alpha x)=d(0,x),\ \forall \alpha \in \mathbb C, |\alpha|=1 \ ?$$ In general, the following holds: $$d(0,\alpha x)\neq |\alpha|d(0,x), \ \alpha \in \mathbb C.$$

• Which examples did you check? – Did Jan 20 '19 at 11:05

For a concrete example take $$\mathbb{C}$$ and let $$\phi\colon\mathbb{C}\to\mathbb{C},\,z\mapsto \operatorname{Re}z+4 \mathrm{i}\operatorname{Im} z.$$ This map is clearly continuous with continuous inverse, thus $$d\colon\mathbb{C}\times \mathbb{C}\to [0,\infty),\,d(z,w)=|\phi(z)-\phi(w)|$$ induces the Eucliean topology on $$\mathbb{C}$$ (which is of course a vector space topology). However $$d(0,2)=2$$, while $$d(0,2i)=8$$.