THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

a) Observe that $$|z+1{|}^2+|z-1{|}^2=(z+1)(\bar{z}+1)+(z-1)(\bar{z}-1)=2|z{|}^2+2=4,$$ hence $|z+1{|}^2-2=2-|z-1{|}^2$, that is, $$(|z+1|-\sqrt{2})(|z+1|+\sqrt{2})=-(|z-1|-\sqrt{2})(|z-1|+\sqrt{2}).$$ Clearly, this implies that $|z+1|-\sqrt{2}$ and $|z-1|-\sqrt{2}$ have opposite signs.

b) Using a), we choose signs such that $|z_k\pm 1|\le \sqrt{2}$. Then $$\sum _{k=1}^{12}|z_k\pm 1|\le 12\sqrt{2}<17,$$ as required.