3. If
$$
\dot{z}=\frac{(1+i)^{2000}(6+2 i)-(1-i)^{1998}(3-i)}{(1+i)^{1996}(23-7 i)+(1-i)^{1994}(10+2 i)} \text {, }
$$

then $|z|=$ . $\qquad$
3.1.
$$
\begin{array}{l}
\text { When }(1+i)^{2}=2 i, (1-i)^{2}=-2 i \text { and } X \in \mathbb{Z}, \\
i^{4 k+1}=i, i^{4 k+2}=-1, i^{4 k+3}=-i, i^{4 k}=1. \\
\text { Therefore, }(1+i)^{2000}=(2 i)^{1000}=2^{1000}, \\
(1-i)^{1998}=(-2 i)^{999}=2^{999} \cdot i, \\
(1+i)^{1996}=(2 i)^{998}=-2^{998}, \\
(1-i)^{1994}=(-2 i)^{997}=-2^{997} i. \\
\text { Then } z=\frac{2^{1000}(6+2 i)-2^{999} i(3-i)}{-2^{998}(23-7 i)-2^{997} i(10+2 i)} \\
=\frac{8(6+2 i)-4 i(3-i)}{-2(23-7 i)-i(10+2 i)} \\
=\frac{48+16 i-12 i-4}{-46+14 i-10 i+2} \\
=\frac{44+4 i}{-44+4 i}=-\frac{11+i}{11-i}. \\
\text { Hence }|z|=\frac{|11+i|}{|11-i|}=1. \\
\end{array}
$$