1. Quadratic equation
$(1-\mathrm{i}) x^{2}+(\lambda+\mathrm{i}) x+(1+\mathrm{i} \lambda)=0$ ( $\mathrm{i}$ 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
$$
\begin{array}{l}
x_{0}{ }^{2}+\lambda x_{0}+1=0 \\
-x_{0}{ }^{2}+x_{0}+\lambda=0
\end{array}
$$
Adding (1) and (2) gives
$$
(\lambda+1)\left(x_{0}+1\right)=0
$$

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

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