For complex number $${z_1} = {x_1} + i{y_1}$$ and $${z_2} = {x_2} + i{y_2},$$ we write $${z_1} \cap {z_2},\,\,if\,\,{x_1} \le {x_2}\,\,and\,\,{y_1} \le {y_2}.$$
Then for all complex numbers $$z\,\,with\,\,1 \cap z,$$ we have $${{1 - z} \over {1 + z}} \cap 0.$$

The complex numbers $$z = x + iy$$ which satisfy the equation $$\,\left| {{{z - 5i} \over {z + 5i}}} \right| = 1$$ lie on

Let the complex number $${{z_1}}$$, $${{z_2}}$$ and $${{z_3}}$$ be the vertices of an equilateral triangle. Let $${{z_0}}$$ be the circumcentre of the triangle. Then prove that $$z_1^2 + z_2^2 + z_3^2 = 3z_0^2$$.

For every integer n > 1, the inequality $${(n!)^{1/n}} < {{n + 1} \over 2}$$ holds.