As well known, $(\mathbb{C}^n, \omega=:\omega_{euclidean})$ is complete . Let $\hat{\omega}$ be a fixed noncomplete metric on $\mathbb{C}^n$.
My question: Can we construct a complete metric on $\mathbb{C}^n$ which coincides with $\hat{\omega}$ on $\mathbb{B}^n$?
Let $\omega_0$ denote the Euclidean metric on ${\mathbb C}^n$ and $\omega$ the given (incomplete) metric. Let $B(0,R)$ denote the open ball centered at the origin, of radius $R$, with respect to the Euclidean metric. Let $\rho$ denote a smooth function on ${\mathbb C}^n$ which equals $1$ on $B(0,1)$, vanishes on the complement to $B(0,2)$ and, overall, takes values in the interval $[0,1]$. (You learn how to construct such functions when you learn about the partition of unity.)
Now, define the following: $$ \omega':= \rho \omega + (1-\rho)\omega_0 $$ I will leave it to you to verify that $\omega'$ is indeed a metric, equal to $\omega$ on $B(0,1)$, equal $\omega_0$ on the complement to $B(0,2)$. You will also verify that $\omega'$ is complete (since it equals the complete metric $\omega_0$ outside of a compact subset.