4. (Mathematics of Intermediate Level, Issue 2, 2004 Olympiad Training Problem) Given two sets of complex numbers $M=\{z \mid z=$ $\left.\cos \alpha+\left(4-\cos ^{2} \alpha\right) \mathrm{i}, \alpha \in \mathbf{R}\right\}, N=\{z \mid z=\cos \beta+(\lambda+\sin \beta) \mathrm{i}, \beta \in \mathbf{R}\}$, when $M \cap N \neq$ $\varnothing$, the range of the real number $\lambda$ is ( ),
A. $[2,5]$
B. $\left[\frac{11}{4}, 5\right]$
C. $\left[\frac{11}{4},+\infty\right)$
D. $(-\infty, 5]$
4. B. Reason: From $M \cap N \neq \varnothing$, we know that the two sets must have the same element. Let $\cos \alpha+\left(4-\cos ^{2} \alpha\right) \mathrm{i}=\cos \beta+(\lambda+\sin \beta) \mathrm{i}$, then $\cos \alpha=\cos \beta$ and $4-\cos ^{2} \alpha=\lambda+\sin \beta$.
Eliminating $\alpha$ gives $\lambda=4-\cos ^{2} \beta-\sin \beta=3+\sin ^{2} \beta-\sin \beta=\frac{11}{4}+\left(\sin \beta-\frac{1}{2}\right)^{2}$.
When $\sin \beta=\frac{1}{2}$ and $\sin \beta=-1$, $\lambda$ respectively attains its minimum value $\frac{11}{4}$ and maximum value 5.