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

$$\begin{gathered} \text{ Let }z=a+ib \\ \Rightarrow \overline{z}=a-ib \\ \begin{array}{l}\text{ So, we have } \\ {z}^{2008}=\overline{z}\\ \therefore |z{|}^{2008}=|\overline{z}|=|z|\\ \Rightarrow \left|z\right|\left[|z{|}^{2007}-1\right]=0\\ \Rightarrow \left|z\right|=0\text{ or }\left|z\right|=1 \\ \text{ if }\left|z\right|=0\text{, then }z=0 \\ \text{ If }\left|z\right|=1\text{, then }{z}^{2009}=z\overline{z}=|z{|}^2=1\text{. } \\ \text{ So, there аге }2009\text{ value of }z\text{. } \\ \text{ Therefore, total number of ordered pairs is }2010.\end{array} \end{gathered}$$