Let $f:B^2(0,1)-\bar{B^2}(0,\frac{1}{2}) \to B^2(0,1)-{0}$ be a homeomorphism. To show that $f$ is not quasiconformal.
I can construct one such homeomorphism explicitly. However, I am not getting why such homeomorphism would not be quasiconformal. Any hints or idea regarding how to proceed ,would be very much helpful.
Quasiconformal mappings almost preserve modulus of annuli, they can distorce it by an amount between $K$ and $1/K$, where $K$ is it qc constant: $$ \frac{1}{K}\mbox{mod}(A)\le \mbox{mod}f(A)\le K \mbox{mod}(A) $$ Modulus of a puncture ball is infinity and the modulus of $B^{2}(0,1)\backslash B^{2}(0,\frac{1}{2})$ is $\frac{1}{2\pi}\log(2)$; so, your homeomorphism cannot be qc.