14. The quadratic equation $(1-i) x^{2}+(\lambda+i) x+(1+i \lambda)=0(i$ being the imaginary unit, $\lambda \in R)$ has two imaginary roots if and only if the range of values for $\lambda$ is $\qquad$ .
14. Consider the opposite of the problem. If the equation has a real root $x_{0}$, then $\left(x_{0}^{2}+\lambda x_{0}+1\right)+i\left(-x_{0}^{2}+x_{0}+\right.$ $\lambda)=0$. This implies $(\lambda+1)\left(x_{0}+1\right)=0$, so $\lambda=-1$ or $x_{0}=-1$.
When $\lambda=-1$, the equation $x_{0}^{2}-x_{0}+1=0$ has no real roots;
When $x_{0}=-1$, $\lambda=2$, i.e., the original equation has real roots if and only if $\lambda=2$.
Therefore, the equation has two complex roots if and only if $\lambda \neq 2$.