4. Given the planar vector $(1,1)$. Then the planar vector $\left(\frac{1-\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right)$ is obtained by transforming the vector $(1,1)$ through $(\quad)$.
(A) Clockwise rotation by $60^{\circ}$
(B) Clockwise rotation by $120^{\circ}$
(C) Counterclockwise rotation by $60^{\circ}$
(D) Counterclockwise rotation by $120^{\circ}$
4. C.

Let the angle between the two vectors be $\theta$.
Then $\cos \theta=\frac{\left(\frac{1-\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right) \cdot(1,1)}{\sqrt{2} \times \sqrt{2}}=\frac{1}{2}$.
Thus, $\theta=60^{\circ}$.
$$
\text { Also, } \frac{1-\sqrt{3}}{2}<0, \theta \in\left[0,180^{\circ}\right] \text {, }
$$

Therefore, option $(\mathrm{C})$ is correct.