2. You might note that $\displaystyle \left| {2a - i} \right| = \left| {2a + i} \right| = \sqrt {4a^2 + 1} \;\& \;\left| {a + i} \right|^4 = \left( {a^2 + 1} \right)^2$.
And so $\displaystyle \left| z \right| = \frac{{\left( {a^2 + 1} \right)^2 }} {{\left( {\sqrt {4a^2 + 1} } \right)^7 }}$.