3. The number of ordered pairs $(a, b)$ that satisfy $(a+b \mathrm{i})^{6}=a-b \mathrm{i}$ (where $\left.a, b \in \mathbf{R}, \mathrm{i}^{2}=-1\right)$ is

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Given $a+b \mathrm{i}=r(\cos \theta+\mathrm{i} \sin \theta)$, then $r^{6}(\cos 6 \theta+\mathrm{i} \sin 6 \theta)=r(\cos \theta-\mathrm{i} \sin \theta)$
$$
\Rightarrow\left\{\begin{array}{l}
r=1, \\
\cos 6 \theta=\cos \theta, \Rightarrow 6 \theta=2 k \pi-\theta \Rightarrow \theta=\frac{2 k \pi}{7}(k=0,1,2,3,4,5,6) . \\
\sin 6 \theta=-\sin \theta
\end{array}\right.
$$

Notice that 0 also satisfies the equation, so there are 8 pairs of $(a, b)$ that meet the requirements.