Number of ordered pairs (a,b) of real numbers such that $(\mathrm{a}+\mathrm{ib}{)}^{2008}=\mathrm{a}-\mathrm{ib}$ holds good, is

Let $\mathrm{z}=\mathrm{a}+\mathrm{ib}\Rightarrow \overline{\mathrm{z}}=\mathrm{a}-\mathrm{ib}$ So we have
  $\begin{array}{l}{ }^{{\mathrm{z}}^{2008}=\overline{\mathrm{z}}}\\ \therefore |\mathrm{z}{|}^{2008}=|\overline{\mathrm{z}}|=|\mathrm{z}|\end{array}$
So, there are 2009 values of z.
Therefore, total number of ordered pairs is 2010.