1. Quadratic equation
$(1-i) x^{2}+(\lambda+i) x+(1+i \lambda)=0 \quad(i$ is the imaginary unit, $\lambda \in R)$ has two imaginary roots if and only if the range of $\lambda$ is $\qquad$ .
1. Suppose the equation has a real root $x_{0}$, then we get
$\left(x_{0}^{2}+\lambda x_{0}+1\right)+i\left(-x_{0}{ }^{2}+x_{0}+\lambda\right)=0$, thus
$$
\begin{array}{l}
x_{0}{ }^{2}+\lambda x_{0}+1=0 \\
-x_{0}{ }^{2}+x_{0}+\lambda=0
\end{array}
$$
(1) + (2) gives
$$
(\lambda+1)\left(x_{0}+1\right)=0
$$

From (1) we get
$$
\left\{\begin{array} { l } 
{ \lambda = - 1 } \\
{ x _ { 0 } { } ^ { 2 } - x _ { 0 } + 1 = 0 }
\end{array} \text { or } \left\{\begin{array}{l}
x_{0}=-1, \\
\lambda=2 .
\end{array}\right.\right.
$$

It can be seen that the original equation has a real root if and only if $\lambda=2$, thus the sufficient and necessary condition for the equation to have two imaginary roots (or no real roots) is $\lambda \neq 2$.